// Numbas version: finer_feedback_settings {"name": "Indices (non-algebraic)", "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], []], "questions": [{"name": "Indices: expanded vs index form (non-algebraic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": ["expanded", "exponent", "exponents", "Exponents", "index", "indices", "power", "powers"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Complete the following without the use of a calculator. Use ^ to signify indices.

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The expression $\\var{latex(expanded)}$ is in expanded form, the same expression in index form is [[0]].

\n

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The expression $\\var{latex(expanded)}$ written in index form is $\\var{base}^\\var{power}$.

\n

The $\\var{base}$ is the base and $\\var{power}$ is the power/exponent/index. The power signifies how many bases are multiplied together.

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You need to use indices. Use ^ to signify indices.

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Nice try!

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The expression $\\var{base}^1$ is normally written as [[0]]

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$\\var{base}^1$ just means there is one $\\var{base}$. So we normally don't write the power. We normally just write $\\var{base}$.

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True or False:

\n


The expression $\\displaystyle\\left(\\frac{\\var{num}}{\\var{den}}\\right)^2$ is equivalent to $\\displaystyle\\frac{\\var{num}^2}{\\var{den}}$.

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The square is acting on the whole fraction:

\n

\\[\\left(\\frac{\\var{num}}{\\var{den}}\\right)^2=\\left(\\frac{\\var{num}}{\\var{den}}\\right)\\times\\left(\\frac{\\var{num}}{\\var{den}}\\right)=\\frac{\\var{num}^2}{\\var{den}^2}\\]

\n
\n

In general $\\left(\\frac{a}{b}\\right)^n=\\frac{a^n}{b^n}$

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True or False:

\n


The expression $\\displaystyle\\left(\\var{num}\\times\\var{den}\\right)^2$ is equivalent to $\\displaystyle\\var{num}^2\\var{den}^2$.

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The square is acting on the whole bracket:

\n

\\[\\left(\\var{num}\\times\\var{den}\\right)^2=\\left(\\var{num}\\times\\var{den}\\right)\\times\\left(\\var{num}\\times\\var{den}\\right)=\\var{num}^2\\times\\var{den}^2\\]

\n
\n

In general $\\left(ab\\right)^n=a^n b^n$

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Evaluate the following, without the use of a calculator:

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For an insight into an index of 0, consider the following table: 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
index form$\\var{base}^3$$\\var{base}^2$$\\var{base}^1$$\\var{base}^0$
result$\\var{base^3}$$\\var{base^2}$$\\var{base}$[[0]]
\n

Notice each time the power decreases by $1$, the result is divided by $\\var{base}$. Using this idea, fill in the rest of the table.

\n

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Each time you reduce the power by 1 you divide the result by the base, that is $\\var{base}$. Following this pattern:

\\[\\var{base}^{0}=\\frac{\\var{base}}{\\var{base}}=1\\]

\n

From this we conclude that any non-zero number to the power of 0 is 1.

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$\\var{int1}^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$\\var{dec1}^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$(\\var{nint1})^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$-(\\var{int2})^0$ = [[0]]

\n

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Here $\\var{int2}$ is to the power of zero and so it evaluates to $1$. However, there is still a negative sign in front, so the answer must be $-1$. 

\n

\n

Note the power of zero is only acting on what is inside the brackets.

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$\\displaystyle\\left(\\simplify{{num1}/{den1}}\\right)^0$ = [[0]]

\n

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Any non-zero number to the power of 0 is 1.

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$\\displaystyle\\frac{\\var{num2}^0}{\\var{den2}}$ = [[0]]

\n

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Note that the power of zero is only acting on the numerator. As such the numerator is equal to $1$ and the denominator remains untouched.

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$(\\var{int1}\\times\\var{int2})^0$ = [[0]]

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Regardless of the result of the multiplication the power of zero will force it to evaluate to 1.

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$\\var{smallint1}\\times\\var{smallint2}^0$ = [[0]]

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The power of zero is only acting on the second number in the product. It is only this number that will become 1. 

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"sumpow1": {"name": "sumpow1", "group": "Ungrouped variables", "definition": "powers1[0][0]+powers1[1][0]+powers1[2][0]", "description": "", "templateType": "anything", "can_override": false}, "sumpow2": {"name": "sumpow2", "group": "Ungrouped variables", "definition": "wild[0]+1", "description": "", "templateType": "anything", "can_override": false}, "sumpow3": {"name": "sumpow3", "group": "Ungrouped variables", "definition": "dec+neg+num/den", "description": "", "templateType": "anything", "can_override": false}, "sumpow4": {"name": "sumpow4", "group": "Ungrouped variables", "definition": "powers3[0][0]+powers3[1][0]", "description": "", "templateType": "anything", "can_override": false}, "neg": {"name": "neg", "group": "Ungrouped variables", "definition": "random(-12..-1)", "description": "", "templateType": "anything", "can_override": false}, "base1": {"name": "base1", "group": "Ungrouped variables", "definition": "primes[0]", "description": "", "templateType": "anything", "can_override": false}, "base2": {"name": "base2", "group": "Ungrouped variables", "definition": "primes[1]", "description": "", "templateType": "anything", "can_override": false}, "num": {"name": "num", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "templateType": "anything", "can_override": false}, "den": {"name": "den", "group": "Ungrouped variables", "definition": "random([10,20,25])", "description": "", "templateType": "anything", "can_override": false}, "wild": {"name": "wild", "group": "Ungrouped variables", "definition": "random([[2,\"two\"],[3,\"three\"],[4,\"four\"],[5,\"five\"],[6,\"six\"]])", "description": "", "templateType": "anything", "can_override": false}, "powers3": {"name": "powers3", "group": "Ungrouped variables", "definition": "shuffle([[2,\"two\"],[3,\"three\"],[4,\"four\"],[5,\"five\"],[6,\"six\"]])[0..2]", "description": "", "templateType": "anything", "can_override": false}, "dec": {"name": "dec", "group": "Ungrouped variables", "definition": "random(1..9)/10", "description": "", "templateType": "anything", "can_override": false}, "powers1": {"name": "powers1", "group": "Ungrouped variables", "definition": "shuffle([[2,\"two\"],[3,\"three\"],[4,\"four\"],[5,\"five\"],[6,\"six\"]])[0..3]", "description": "

2..6

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$\\var{base1}^\\var{powers1[0][0]}\\times \\var{base1}^\\var{powers1[1][0]} \\times \\var{base1}^\\var{powers1[2][0]}$ = [[0]]

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Recall:

\n\n

So in total how many $\\var{base1}$s are there multiplied together? 

\n

Well, $\\var{powers1[0][0]}+\\var{powers1[1][0]}+\\var{powers1[2][0]}=\\var{sumpow1}$. And so our answer is $\\var{base1}^\\var{sumpow1}$.

\n

Note, in general $a^ba^c=a^{b+c}$.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\var{base2}^\\var{powers2[0][0]}\\times \\var{base2}^\\var{powers2[1][0]} \\times \\var{base2}^\\var{powers2[2][0]}$ = [[0]]

\n

\n

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Recall:

\n\n

So in total how many $\\var{base2}$s are there multiplied together? 

\n

Well, $0+1+\\var{wild[0]}=\\var{sumpow2}$. And so our answer is $\\var{base2}^\\var{sumpow2}$.

\n

Note, in general $a^ba^c=a^{b+c}$, $a^0=1$ and $a^1=a$.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n


$\\displaystyle\\var{base2}^\\var{dec}\\times \\var{base2}^\\var{neg} \\times \\var{base2}^{\\var{num}/\\var{den}}$ = [[0]]

\n

\n

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

\n

\n

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Since the bases are all the same ($\\var{base2}$) and we are multiplying, we can simply add the powers. 

\n

You could convert the fraction to a decimal and then add them all. Or you could add them all as fractions. 

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers:

\n

$\\var{base3}^\\var{powers3[0][0]}\\times \\var{base1}^\\var{wild[0]} \\times \\var{base3}^\\var{powers3[1][0]}$ = [[0]]

\n

Note: use * for multiplication.

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It is important to note that the bases are different! We can only add the powers if the bases are the same. 

\n

Recall:

\n\n

So in total what do we have?

\n

 $\\var{base1}^\\var{wild[0]}\\var{base3}^\\var{sumpow4}$.

\n

Note we would type {base1}^{wild[0]}*{base3}^{sumpow4}.

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Use ^ for powers. Input your answer in index form.

"}, "notallowed": {"strings": ["^0", "^1"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Is the following statement true or false?

\n

$\\var{base1}\\times\\var{base2}^\\var{powers1[0][0]} = \\var{base1*base2}^\\var{powers1[0][0]}$

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It is important to note that the bases are different! Index laws only can be applied if the bases are the same (or can be made the same).

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Is the following statement true or false?

\n

$\\var{base1}\\times\\var{base2}^\\var{powers1[0][0]} = \\var{base1*base2}^\\var{powers1[0][0]+1}$

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It is important to note that the bases are different! Index laws only can be applied if the bases are the same (or can be made the same). We can only add the powers if the bases are the same. 

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Use ^ to signify powers.

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2..6

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For an insight into negative indices, consider the following table: 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
index form$\\var{base}^3$$\\var{base}^2$$\\var{base}^1$$\\var{base}^0$$\\var{base}^{-1}$$\\var{base}^{-2}$
result$\\var{base^3}$$\\var{base^2}$$\\var{base}$$1$[[0]][[1]]
\n

Notice each time the power decreases by $1$, the result is divided by $\\var{base}$. Using this idea, and / for division, fill in the rest of the table.

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Each time you reduce the power by 1 you divide the result by the base, that is $\\var{base}$. Following this pattern:

\\[\\var{base}^{-1}=\\frac{1}{\\var{base}}\\]

\n

and 

\n

\\[\\var{base}^{-2}=\\frac{1}{\\var{base}}\\div\\var{base}=\\frac{1}{\\var{base}^2}\\]

\n

\n
\n

\n

Note you could input the second expression as $1/\\var{base}/\\var{base}$ or $1/\\var{base}\\wedge2$ or $1/\\var{base^2}$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "1/{base}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.0000001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "musthave": {"strings": ["/"], "showStrings": false, "partialCredit": 0, "message": ""}, "notallowed": {"strings": ["^0", "^1", "^-"], "showStrings": false, "partialCredit": 0, "message": "

Don't use negative powers here in this table.

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Don't use negative powers here in this table.

"}, "valuegenerators": []}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The fraction $\\displaystyle\\frac{1}{\\var{base1}^\\var{power1}}$ can be written using a negative index as [[0]].

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Notice:
\\[1=\\var{base1}^0=\\var{base1}^{\\var{power1}-\\var{power1}}=\\var{base1}^{\\var{power1}} \\var{base1}^{-\\var{power1}}\\]

\n

 That is, we have

\n

\\[1=\\var{base1}^{\\var{power1}} \\var{base1}^{-\\var{power1}}\\]

\n

by dividing both sides of this equation by $\\var{base1}^{\\var{power1}}$ we get

\n

\\[\\frac{1}{\\var{base1}^{\\var{power1}}}=\\var{base1}^{-\\var{power1}}\\]

\n
\n

In general, we have $\\displaystyle\\frac{1}{a^c}=a^{-c}$.

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Use ^ for powers. Input your answer in index form.

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Write your answer with a negative index.

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The expression $\\var{base2}^{\\var{power2}}$ can be written without a negative index as the fraction  [[0]].

\n

\n

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Notice:
\\[1=\\var{base2}^0=\\var{base2}^{\\var{power2}+\\var{-power2}}=\\var{base2}^{\\var{power2}} \\var{base2}^{\\var{-power2}}\\]

\n

 That is, we have

\n

\\[1=\\var{base2}^{\\var{power2}} \\var{base2}^{\\var{-power2}}\\]

\n

by dividing both sides of this equation by $\\var{base2}^{\\var{-power2}}$ we get

\n

\\[\\frac{1}{\\var{base2}^{\\var{-power2}}}=\\var{base2}^{\\var{power2}}\\]

\n
\n

In general, we have $\\displaystyle\\frac{1}{a^c}=a^{-c}$.

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Use ^ for powers. Input your answer in index form.

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The expression $\\displaystyle\\frac{1}{\\var{base3}^{\\var{-power3}}}$ can be written without a negative index and without the use of a fraction.

\n

In particular, $\\displaystyle\\frac{1}{\\var{base3}^{\\var{-power3}}}$ in index form would be [[0]].

\n

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You can think of the negative power as forcing the term to the other part of the fraction (if it was on top it goes to the bottom and if it was on the bottom it goes to the top) except now it has a positive power.

\n
\n

We can see this by using our rules for dividing fractions:

\n

\\[\\frac{1}{\\var{base3}^\\var{-power3}}=\\frac{1}{\\left(\\frac{1}{\\var{base3}^\\var{power3}}\\right)}=1\\div{\\left(\\frac{1}{\\var{base3}^\\var{power3}}\\right)}=1\\times \\left(\\frac{{\\var{base3}^\\var{power3}}}{1}\\right)=\\var{base3}^\\var{power3}\\]

\n
\n

In general, we have $\\displaystyle\\frac{1}{a^c}=a^{-c}$ and $\\displaystyle\\frac{1}{a^{-c}}=a^{c}$.

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Use index form.

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Don't use fractions or negative indices.

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers. Use negative powers if necessary.

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2..6

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$\\var{base1}^\\var{powers1[0][0]}\\div \\var{base1}^\\var{powers1[1][0]}$ = [[0]]

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Write the division as a fraction and cancel common factors or use the index law $\\displaystyle\\frac{a^b}{a^c}=a^{b-c}$. 

\n

Note, negative indices mean you divided by more than you had, for example, $\\displaystyle \\frac{1}{12^3}$ can be written as $12^{-3}$.

\n

For our particular question we have, $\\var{base1}^\\var{powers1[0][0]}\\div \\var{base1}^\\var{powers1[1][0]}=\\frac{\\var{base1}^\\var{powers1[0][0]}}{\\var{base1}^\\var{powers1[1][0]}}$ is equal to $\\var{base1}^{\\var{powers1[0][0]}-\\var{powers1[1][0]}}=\\var{base1}^{\\var{diffpow1}}$.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle\\frac{\\var{base2}^\\var{powers2[0]}\\times\\var{base2}}{\\var{base2}^\\var{powers2[1]} \\times \\var{base2}^\\var{powers2[2]}}$ = [[0]]

\n

\n

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Cancel common factors or use the index laws, e.g. $\\displaystyle\\frac{a^b}{a^c}=a^{b-c}$ and $a^b\\times a^c = a^{b+c}$.

\n

Note, negative indices mean you divided by more than you had, for example, $\\displaystyle \\frac{1}{12^3}$ can be written as $12^{-3}$.

\n

For $\\frac{\\var{base2}^\\var{powers2[0]}\\times\\var{base2}}{\\var{base2}^\\var{powers2[1]} \\times \\var{base2}^\\var{powers2[2]}}$ we will simplify the numerator and denominator by using $a^b\\times a^c = a^{b+c}$ and end up with $\\frac{\\var{base2}^\\var{powers2[0]+1}}{\\var{base2}^\\var{powers2[1]+powers2[2]}}$.  Then we will use $\\frac{a^b}{a^c}=a^{b-c}$ to finally get $\\var{base2}^\\var{powers2[0]+1-powers2[1]-powers2[2]}$.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n

$\\displaystyle\\frac{\\var{base2}^\\var{ndec}}{\\var{base2}^\\var{neg}}$ = [[0]]

\n

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

\n

\n

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Since the bases are all the same ($\\var{base2}$) and we are dividing, we can subtract the powers since in general, we have $\\displaystyle\\frac{a^b}{a^c}=a^{b-c}$. Also recall that subtracting a negative is equivalent to adding a positive.

\n

$\\begin{align}\\frac{\\var{base2}^\\var{ndec}}{\\var{base2}^\\var{neg}}&=\\var{base2}^{\\var{ndec}-\\var{neg}}\\\\
&=\\var{base2}^{\\var{ndec}+\\var{abs(neg)}}\\\\&=\\var{base2}^{\\var{ndec+abs(neg)}}\\end{align}$

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers. Use negative powers if necessary.

\n

$\\var{base3}^\\var{powers3[0][0]}\\times \\var{base1}^\\var{powers3[1][0]} \\div \\var{base3}^\\var{powers3[2][0]}$ = [[0]]

\n

Note: use * for multiplication.

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It is important to note that the bases are different! Index laws can only be applied if the bases are the same (or can be made the same). Because of this, we deal with the different bases separately. 

\n

Notice the first part of the expression can not be simplified using index laws.

\n

$\\var{base3}^\\var{powers3[0][0]}\\times \\var{base1}^\\var{powers3[1][0]}$ 

\n

However, with the division we can do some simplification. We can either:

\n
    \n
  1. write it as a fraction and cancel the common factor of $\\var{base3}^\\var{minpow4}$ from the top and bottom:
    \\[\\frac{\\var{base3}^\\var{powers3[0][0]}\\times \\var{base1}^\\var{powers3[1][0]}}{ \\var{base3}^\\var{powers3[2][0]}}=\\frac{\\var{base3}^\\var{powers3[0][0]-minpow4}\\times \\var{base1}^\\var{powers3[1][0]}}{ \\var{base3}^\\var{powers3[2][0]-minpow4}}=\\var{base3}^\\var{diffpow4}\\var{base1}^\\var{powers3[1][0]}\\] 
  2. \n
  3. Subtract the powers, \"top power minus the bottom power\" for the terms with the same base:
    \\[\\frac{\\var{base3}^\\var{powers3[0][0]}\\times \\var{base1}^\\var{powers3[1][0]}}{ \\var{base3}^\\var{powers3[2][0]}}=\\var{base3}^\\var{diffpow4}\\var{base1}^\\var{powers3[1][0]}\\] 
  4. \n
"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{base1}^{powers3[1][0]}*{base3}^{diffpow4}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "musthave": {"strings": ["^", "*"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Is the following statement true or false?

\n

$\\displaystyle\\frac{\\var{2*base2}}{\\var{base2}^\\var{powers1[0][0]}} = \\var{2}^\\var{-powers1[0][0]}$

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It is important to note that the bases are different! Index laws only can be applied if the bases are the same (or can be made the same).

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Is the following statement true or false?

\n

$\\displaystyle\\frac{\\var{2*base2}}{\\var{base2}^\\var{powers1[0][0]}} = \\var{2}^\\var{1-powers1[0][0]}$

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It is important to note that the bases are different! Index laws only can be applied if the bases are the same (or can be made the same). We can only add (or subtract) the powers if the bases are the same. 

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Is the following statement true or false?

\n

$\\displaystyle\\frac{\\var{2*base2}}{\\var{base2}^\\var{powers1[0][0]}} = 2\\times\\var{base2}^\\var{1-powers1[0][0]}$

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It is important to note that the bases are different! Index laws can only be applied if the bases are the same (or can be made the same). We can only add (or subtract) the powers if the bases are the same. 

\n

Note in this question we can make the bases the same.

\n

\\[\\frac{\\var{2*base2}}{\\var{base2}^\\var{powers1[0][0]}} = \\frac{2\\times\\var{base2}}{\\var{base2}^\\var{powers1[0][0]}} = 2\\times\\var{base2}^\\var{1-powers1[0][0]}\\]

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"num": {"name": "num", "group": "Ungrouped variables", "definition": "random(-9..-1)", "description": "", "templateType": "anything", "can_override": false}, "nint": {"name": "nint", "group": "Ungrouped variables", "definition": "random(-12..-2)", "description": "", "templateType": "anything", "can_override": false}, "base1": {"name": "base1", "group": "Ungrouped variables", "definition": "random(2,3,5,7)", "description": "", "templateType": "anything", "can_override": false}, "ndec": {"name": "ndec", "group": "Ungrouped variables", "definition": "random(-0.9..-0.1#0.1)", "description": "", "templateType": "anything", "can_override": false}, "den": {"name": "den", "group": "Ungrouped variables", "definition": "random(10,20,25)", "description": "", "templateType": "anything", "can_override": false}, "prodpow4": {"name": "prodpow4", "group": "Ungrouped variables", "definition": "nint*ndec*num/den", "description": "", "templateType": "anything", "can_override": false}, "base2": {"name": "base2", "group": "Ungrouped variables", "definition": "random([2,3,5,7] except base1)", "description": "", "templateType": "anything", "can_override": false}, "base3": {"name": "base3", "group": "Ungrouped variables", "definition": "primes[0]", "description": "", "templateType": "anything", "can_override": false}, "prodpow1": {"name": "prodpow1", "group": "Ungrouped variables", "definition": "2*powers1", "description": "", "templateType": "anything", "can_override": false}, "prodpow3": {"name": "prodpow3", "group": "Ungrouped variables", "definition": "powers3[0][0]*powers3[1][0]*powers3[2][0]", "description": "", "templateType": "anything", "can_override": false}, "prodpow2": {"name": "prodpow2", "group": "Ungrouped variables", "definition": "powers2[0][0]*powers2[1][0]", "description": "", "templateType": "anything", "can_override": false}, "powers1": {"name": "powers1", "group": "Ungrouped variables", "definition": "random(3, 4, 10, 11, 12)", "description": "

2..6

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$\\left(\\var{base1}^\\var{powers1}\\right)^2$ = [[0]]

\n

Hint: Your answer should be written as $\\var{base1}$ to some power.

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Interpret the powers in expanded form and then write the result in index form, or use the index law $\\displaystyle(a^b)^c=a^{bc}$.

\n

\\begin{align}\\left(\\var{base1}^\\var{powers1}\\right)^2&=\\var{base1}^{\\var{powers1}\\times 2}\\\\
&=\\var{base1}^\\var{2*powers1}\\end{align}

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Your answer is longer than necessary.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle(\\var{base2}^\\var{powers2[0][0]})^\\var{powers2[1][0]}$ = [[0]]

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Interpret the powers in expanded form and then write the result in index form, or use the index law $\\displaystyle(a^b)^c=a^{bc}$.

\n

\\begin{align}(\\var{base2}^\\var{powers2[0][0]})^\\var{powers2[1][0]}&=\\var{base2}^{\\var{powers2[0][0]}\\times\\var{powers2[1][0]}}\\\\
&=\\var{base1}^\\var{powers2[0][0]*powers2[1][0]}\\end{align}

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Your answer is longer than necessary.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n


$\\displaystyle\\left(\\left(\\var{base3}^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}$ = [[0]]

\n

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Interpret the powers in expanded form and then write the result in index form, or use the index law $\\displaystyle(a^b)^c=a^{bc}$. 

\n

We could work from the inside to the outside:

\n

\\begin{align}\\left(\\left(\\var{base3}^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}&=\\left(\\var{base3}^{\\var{powers3[0][0]}\\times \\var{powers3[1][0]}}\\right)^\\var{powers3[2][0]}\\\\
&=\\left(\\var{base3}^{\\var{powers3[0][0]*powers3[1][0]}}\\right)^\\var{powers3[2][0]}\\\\
&=\\var{base3}^{\\var{powers3[0][0]*powers3[1][0]}\\times\\var{powers3[2][0]}}\\\\
&=\\var{base3}^{\\var{powers3[0][0]*powers3[1][0]*powers3[2][0]}}\\\\
\\end{align}

\n

We could also work from the outside to the inside:

\n

\\begin{align}\\left(\\left(\\var{base3}^\\var{powers3[0][0]}\\right)^\\var{powers3[1][0]}\\right)^\\var{powers3[2][0]}&=\\left(\\var{base3}^\\var{powers3[0][0]}\\right)^{\\var{powers3[1][0]}\\times\\var{powers3[2][0]}}\\\\
&=\\left(\\var{base3}^\\var{powers3[0][0]}\\right)^{\\var{powers3[1][0]*powers3[2][0]}}\\\\
&=\\var{base3}^{\\var{powers3[0][0]}\\times\\var{powers3[1][0]*powers3[2][0]}}\\\\
&=\\var{base3}^{\\var{powers3[0][0]*powers3[1][0]*powers3[2][0]}}\\\\
\\end{align}

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Your answer is longer than necessary.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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$\\displaystyle\\left(\\left(\\var{base4}^\\var{nint}\\right)^\\var{ndec}\\right)^{\\var{num}/\\var{den}}$ = [[0]]

\n

\n

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

\n

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Working from the outside to the inside, using the index law $\\displaystyle(a^b)^c=a^{bc}$: 

\n

\\begin{align}\\left(\\left(\\var{base4}^\\var{nint}\\right)^\\var{ndec}\\right)^{\\var{num}/\\var{den}}
&=\\left(\\var{base4}^\\var{nint}\\right)^{\\var{ndec}\\times\\frac{\\var{num}}{\\var{den}}}\\\\
&=\\left(\\var{base4}^\\var{nint}\\right)^{\\var[fractionnumbers]{ndec*num/den}}\\\\
&=\\var{base4}^{\\var{nint}\\times\\var[fractionnumbers]{ndec*num/den}}\\\\
&=\\var{base4}^{\\var[fractionnumbers]{nint*ndec*num/den}}
\\end{align}

\n

Note the answer can be written using a fractional power or a decimal power but a fractional power is more commonly used.

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Your answer is longer than it needs to be.

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Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

"}, "mustmatchpattern": {"pattern": "$n^$n", "partialCredit": 0, "message": "", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Indices: fractional powers (non-algebraic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": ["exponent", "exponents", "Exponents", "fractional", "index", "index laws", "Index Laws", "indices", "power", "powers", "rational", "roots"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"cube": {"name": "cube", "group": "Ungrouped variables", "definition": "random([[8,4],[27,9]])", "description": "", "templateType": "anything", "can_override": false}, "root1": {"name": "root1", "group": "Ungrouped variables", "definition": "random(4..12)", "description": "", "templateType": "anything", "can_override": false}, "power2": {"name": "power2", "group": "Ungrouped variables", "definition": "random(2..12 except root2)", "description": "", "templateType": "anything", "can_override": false}, "base1": {"name": "base1", "group": "Ungrouped variables", "definition": "random(2,3,5,7,11,13,17,19)", "description": "", "templateType": "anything", "can_override": false}, "base2": {"name": "base2", "group": "Ungrouped variables", "definition": "random([2,3,5,7] except base1)", "description": "", "templateType": "anything", "can_override": false}, "square": {"name": "square", "group": "Ungrouped variables", "definition": "random([[4,2],[9,3],[16,4],[25,5],[36,6],[49,7],[64,8],[81,9],[121,11],[144,12]])", "description": "", "templateType": "anything", "can_override": false}, "base4": {"name": "base4", "group": "Ungrouped variables", "definition": "primes[1]", "description": "", "templateType": "anything", "can_override": false}, "zeros": {"name": "zeros", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything", "can_override": false}, "primes": {"name": "primes", "group": "Ungrouped variables", "definition": "shuffle([11,13,17,19])[0..2] ", "description": "", "templateType": "anything", "can_override": false}, "root2": {"name": "root2", "group": "Ungrouped variables", "definition": "random(2..12)", "description": "", "templateType": "anything", "can_override": false}, "power3": {"name": "power3", "group": "Ungrouped variables", "definition": "random(2..12 except [root2,root3])", "description": "", "templateType": "anything", "can_override": false}, "onezeros": {"name": "onezeros", "group": "Ungrouped variables", "definition": "10^zeros", "description": "", "templateType": "anything", "can_override": false}, "root3": {"name": "root3", "group": "Ungrouped variables", "definition": "random(2..12)", "description": "", "templateType": "anything", "can_override": false}, "base3": {"name": "base3", "group": "Ungrouped variables", "definition": "primes[0]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["primes", "base1", "base2", "base3", "base4", "root1", "root2", "power2", "root3", "power3", "square", "cube", "zeros", "onezeros"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

By using the definition of the square root you should see that $(\\sqrt{\\var{base1}})^2=\\var{base1}$.

\n

By using index laws you should see that $(\\var{base1}^{1/2})^2=\\var{base1}$.

\n

The above equations imply that $\\sqrt{\\var{base1}}$ can also be written as  [[0]].

\n

\n

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

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Given \\[(\\sqrt{\\var{base1}})^2=\\var{base1}=(\\var{base1}^{1/2})^2\\]

\n

we can say \\[\\sqrt{\\var{base1}}=\\var{base1}^{1/2}\\]

\n

Which we would type in as $\\var{base1}\\wedge(1/2)$.

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Your answer is longer than necessary.

"}, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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By using the definition of the cube root you should see that $(\\sqrt[3]{\\var{base2}})^3=\\var{base2}$.

\n

By using index laws you should see that $(\\var{base2}^{1/3})^3=\\var{base2}$.

\n

The above equations imply that $\\sqrt[3]{\\var{base2}}$ can also be written as  [[0]].

\n

\n

Note: If you want to use a fraction as a power you should use brackets to surround your power, for example, type 12^(2/3) for $12^\\frac{2}{3}$.

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Given \\[(\\sqrt[3]{\\var{base2}})^3=\\var{base2}=(\\var{base2}^{1/3})^3\\]

\n

we can say \\[\\sqrt[3]{\\var{base2}}=\\var{base2}^{1/3}\\]

\n

Which we would type in as $\\var{base2}\\wedge(1/3)$.

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Your answer is longer than necessary.

"}, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n


$\\sqrt[\\var{root1}]{\\var{base3}}$ = [[0]]

\n

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In general, we have $\\sqrt[n]{a}=a^{\\frac{1}{n}}$.

\n

In particular, $\\sqrt[\\var{root1}]{\\var{base3}}=\\var{base3}^\\frac{1}{\\var{root1}}$.

\n

 

\n

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{base3}^(1/{root1})", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.00001", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "maxlength": {"length": "11", "partialCredit": 0, "message": "

Your answer is longer than necessary.

"}, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

"}, "notallowed": {"strings": ["^0", "sqrt", "root"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n

$\\displaystyle\\left(\\sqrt[\\var{root2}]{\\var{base4}}\\right)^\\var{power2}$ = [[0]]

\n

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Convert the root to a fractional power and then use the index laws to deal with the two different powers.

\n

That is, \\begin{align}\\left(\\sqrt[\\var{root2}]{\\var{base4}}\\right)^\\var{power2}
&=\\left(\\var{base4}^{\\frac{1}{\\var{root2}}}\\right)^\\var{power2}\\\\
&=\\var{base4}^{\\frac{1}{\\var{root2}}\\times\\var{power2}}\\\\
&=\\var{base4}^{\\var[fractionnumbers]{power2/root2}}.\\end{align}

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Your answer is longer than necessary.

"}, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

"}, "notallowed": {"strings": ["^0", "sqrt", "root"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

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Use the same approach you used in the above questions to simplify the following in index form.

\n

$\\sqrt[\\var{root3}]{\\var{base1}^\\var{power3}}$ = [[0]]

\n

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Convert the root to a fractional power and then use the index laws to deal with the two different powers.

\n

That is, \\begin{align}\\left(\\sqrt[\\var{root3}]{\\var{base1}}\\right)^\\var{power3}
&=\\left(\\var{base1}^{\\frac{1}{\\var{root3}}}\\right)^\\var{power3}\\\\
&=\\var{base1}^{\\frac{1}{\\var{root3}}\\times\\var{power3}}\\\\
&=\\var{base1}^{\\var[fractionnumbers]{power3/root3}}.\\end{align}

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Your answer is longer than necessary.

"}, "musthave": {"strings": ["^"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

"}, "notallowed": {"strings": ["^0", "sqrt", "root"], "showStrings": false, "partialCredit": 0, "message": "

Use ^ for powers. Input your answer in index form.

"}, "mustmatchpattern": {"pattern": "?^?", "partialCredit": 0, "message": "", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": []}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

By interpreting the denominator of the fractional power as an nth root, determine the value of the following:

\n


$\\var{square[0]}^{\\frac{1}{2}}$ = [[0]]

\n

$\\var{cube[0]}^{\\frac{2}{3}}$ = [[1]]

\n

$\\var{onezeros}^{\\frac{1}{\\var{zeros}}}$ = [[2]]

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Convert the denominator of the fractional power to a root and use your multiplication knowledge.

\n

That is,

\n

\\begin{align}\\var{square[0]}^{\\frac{1}{2}}
&=\\sqrt{\\var{square[0]}}\\\\
&=\\var{square[1]}&(\\text{since } \\var{square[1]}^2=\\var{square[0]})\\end{align}

\n

and
\\begin{align}\\var{cube[0]}^{\\frac{2}{3}}
&=\\left(\\var{cube[0]}^{\\frac{1}{3}}\\right)^2\\\\
&=\\left(\\sqrt[3]{\\var{cube[0]}}\\right)^2\\\\
&={\\var{sqrt(cube[1])}}^2&(\\text{since } \\var{sqrt(cube[1])}^3=\\var{cube[0]})\\\\
&=\\var{cube[1]},\\end{align}

\n

and finally

\n

\\begin{align}\\var{onezeros}^{\\frac{1}{\\var{zeros}}}
&=\\sqrt[\\var{zeros}]{\\var{onezeros}}\\\\
&=10&(\\text{since } 10^\\var{zeros}=\\var{onezeros}).\\end{align}

\n

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Simplify the following without the use of a calculator. Write your answer in index form using ^ to signify powers.

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If you would like a challenge, then here is a question for you. Use your index laws rather than your calculator to see how this simplifies:

\n

$\\displaystyle \\frac{\\sqrt[\\var{root}]{\\var{a}^\\var{n}\\times\\var{b}\\times\\var{a}^\\var{root-n}}}{\\var{c}^\\var{-m}\\times\\var{a}\\times\\var{b}^0\\times\\var{b}^{1/\\var{root}}}\\times \\left(\\frac{\\var{a}^\\var{p}\\times\\var{b}}{\\var{c}}\\right)^\\var{m}+\\var{ans}(\\var{a}^\\var{root-n}\\times\\var{b}^\\var{m+root}\\times\\var{c}^\\var{n+m})^0-\\frac{\\var{a}^\\var{p*m-n}\\times\\var{a}^\\var{n}}{\\var{b}^\\var{-m}}$ = [[0]]

\n

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The order shown is simply one order in which to do things, you also don't need to use as many lines as are shown here. There are only three terms in this expression. We will simplify them individually and then combine them.

\n

The middle term $\\var{ans}(\\var{a}^\\var{root-n}\\times\\var{b}^\\var{m+root}\\times\\var{c}^\\var{n+m})^0$ has a power of zero acting on a bracket so this can be reduced to simply $\\var{ans}\\times 1=\\var{ans}$.

\n

The third term is slightly more complicated in that we will need to add indices (because we are multiplying expressions with the same base) and deal with negative indices:

\n

\\begin{align}&\\frac{\\var{a}^\\var{p*m-n}\\times\\var{a}^\\var{n}}{\\var{b}^\\var{-m}}\\\\
&=\\frac{\\var{a}^\\color{red}{\\var{p*m}}}{\\var{b}^\\var{-m}}\\\\
&=\\var{a}^\\var{p*m}\\times\\color{red}{\\var{b}^\\var{m}}\\end{align}

\n

The first term is more complicated and we will use all of our index laws to simplify it:
\\begin{align}&\\frac{\\sqrt[\\var{root}]{\\var{a}^\\var{n}\\times\\var{b}\\times\\var{a}^\\var{root-n}}}{\\var{c}^\\var{-m}\\times\\var{a}\\times\\var{b}^0\\times\\var{b}^{1/\\var{root}}}\\times \\left(\\frac{\\var{a}^\\var{p}\\times\\var{b}}{\\var{c}}\\right)^\\var{m}\\\\
&=\\frac{\\left(\\var{a}^\\var{n}\\times\\var{b}\\times\\var{a}^\\var{root-n}\\right)^{\\color{red}{1/\\var{root}}}}{\\var{c}^\\var{-m}\\times\\var{a}\\times \\color{red}{1}\\times\\var{b}^{1/\\var{root}}}\\times \\left(\\frac{\\var{a}^\\color{red}{\\var{p*m}}\\times\\var{b}^\\color{red}{\\var{m}}}{\\var{c}^\\color{red}{\\var{m}}}\\right)\\\\
&=\\frac{\\left(\\var{a}^\\color{red}{\\var{root}}\\times\\var{b}\\right)^{1/\\var{root}}\\color{red}{\\times}\\var{a}^\\var{p*m}\\times\\var{b}^\\var{m}}{\\var{c}^\\var{-m}\\times\\var{a}\\times\\var{b}^{1/\\var{root}}\\color{red}{\\times} \\var{c}^\\var{m}}\\\\
&=\\frac{\\var{a}^\\color{red}{1}\\times\\var{b}^\\color{red}{1/\\var{root}}\\color{red}{\\times}\\var{a}^\\var{p*m}\\times\\var{b}^\\var{m}}{\\var{c}^\\color{red}{\\var{0}}\\times\\var{a}\\times\\var{b}^{1/\\var{root}}}\\\\
&=\\frac{\\var{a}^\\color{red}{\\var{1+p*m}}\\times\\var{b}^\\color{red}{\\var{m}+1/\\var{root}}}{\\color{red}{1}\\times\\var{a}\\times\\var{b}^{1/\\var{root}}}\\\\
&=\\var{a}^\\color{red}{\\var{p*m}}\\times\\var{b}^\\color{red}{\\var{m}}\\\\
\\end{align}

\n

This is the same as the third term.

\n

So far we have reduced our original expression to $\\var{a}^\\color{red}{\\var{p*m}}\\times\\var{b}^\\color{red}{\\var{m}}+\\var{ans}-\\var{a}^\\color{red}{\\var{p*m}}\\times\\var{b}^\\color{red}{\\var{m}}$ but this is simply $\\var{ans}$.

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Your answer is longer than necessary.

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