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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]].

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For example, $5\\times 5\\times 5\\times 5$ is written as $5^4$ in index form.

In the above example, 5 is the base and 4 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]]

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

True or False:

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:

\\[\\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}\\]

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

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True or False:

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

The square is acting on the whole bracket:

\\[\\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\\]

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

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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]]

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.

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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\\]

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

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$\\var{int1}^0$ = [[0]]

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

$\\var{dec1}^0$ = [[0]]

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

$(\\var{nint1})^0$ = [[0]]

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

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

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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 $-1$.

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

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

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

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

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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.

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

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

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

Evaluate the following, without the use of a calculator:

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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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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]]

Recall:

$\\var{base1}^\\var{powers1[0][0]}$ means $\\var{powers1[0][1]}$ $\\var{base1}$s all multiplied together.
$\\var{base1}^\\var{powers1[1][0]}$ means $\\var{powers1[1][1]}$ $\\var{base1}$s all multiplied together.
$\\var{base1}^\\var{powers1[2][0]}$ means $\\var{powers1[2][1]}$ $\\var{base1}$s all multiplied together.

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

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

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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.

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

Recall:

$\\var{base2}^0$ means no $\\var{base2}$s all multiplied together. (As you know, $\\var{base2}^0=1$).
$\\var{base2}^1$ means just one $\\var{base2}$. (As you know, $\\var{base2}^1=\\var{base2}$)
$\\var{base2}^\\var{wild[0]}$ means $\\var{wild[1]}$ $\\var{base2}$s all multiplied together.

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

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

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.

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

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

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 $12^\\frac{2}{3}$.

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

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.

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

Note: use * for multiplication.

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

Recall:

$\\var{base3}^\\var{powers3[0][0]}$ means $\\var{powers3[0][1]}$ $\\var{base3}$s all multiplied together.
$\\var{base3}^\\var{powers3[1][0]}$ means $\\var{powers3[1][1]}$ $\\var{base3}$s all multiplied together.
$\\var{base1}^\\var{wild[0]}$ means $\\var{wild[1]}$ $\\var{base1}$s all multiplied together.

So in total what do we have?

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

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

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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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Is the following statement true or false?

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

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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True

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False

Is the following statement true or false?

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

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.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"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}, "base3": {"name": "base3", "group": "Ungrouped variables", "definition": "primes[2]", "description": "", "templateType": "anything", "can_override": false}, "base": {"name": "base", "group": "Ungrouped variables", "definition": "random(2,3,5,10)", "description": "", "templateType": "anything", "can_override": false}, "primes": {"name": "primes", "group": "Ungrouped variables", "definition": "shuffle([2,3,5,7,11,13,17])[0..3] ", "description": "", "templateType": "anything", "can_override": false}, "power3": {"name": "power3", "group": "Ungrouped variables", "definition": "random(2..6)", "description": "", "templateType": "anything", "can_override": false}, "power2": {"name": "power2", "group": "Ungrouped variables", "definition": "random(-6..-2)", "description": "", "templateType": "anything", "can_override": false}, "power1": {"name": "power1", "group": "Ungrouped variables", "definition": "random(2..6)", "description": "

2..6

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["primes", "base1", "power1", "base2", "power2", "base3", "power3", "base"], "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": "

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]]

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.

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}}\\]

and

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

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

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Don't use negative powers here in this table.

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Don't use negative powers here in this table.

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

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

That is, we have

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

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

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

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.

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

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

That is, we have

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

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

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

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

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

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

The simplest way to write $\\displaystyle\\frac{1}{\\var{base3}^{\\var{-power3}}}$ in index form would be [[0]]

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.

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

\\[\\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}\\]

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.

", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"minpow4": {"name": "minpow4", "group": "Ungrouped variables", "definition": "min(powers3[0][0],powers3[2][0])", "description": "", "templateType": "anything", "can_override": false}, "diffpow2": {"name": "diffpow2", "group": "Ungrouped variables", "definition": "powers2[0]+1-powers2[1]-powers2[2]", "description": "", "templateType": "anything", "can_override": false}, "neg": {"name": "neg", "group": "Ungrouped variables", "definition": "random(-12..-1)", "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}, "diffpow1": {"name": "diffpow1", "group": "Ungrouped variables", "definition": "powers1[0][0]-powers1[1][0]", "description": "", "templateType": "anything", "can_override": false}, "diffpow3": {"name": "diffpow3", "group": "Ungrouped variables", "definition": "ndec-2*neg/2", "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}, "base3": {"name": "base3", "group": "Ungrouped variables", "definition": "primes[2]", "description": "", "templateType": "anything", "can_override": false}, "diffpow4": {"name": "diffpow4", "group": "Ungrouped variables", "definition": "powers3[0][0]-powers3[2][0]", "description": "", "templateType": "anything", "can_override": false}, "powers1": {"name": "powers1", "group": "Ungrouped variables", "definition": "shuffle([[2,\"two\"],[8,\"eight\"],[4,\"four\"],[10,\"ten\"],[6,\"six\"]])[0..2]", "description": "

2..6

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

Write the division as a fraction and cancel common factors.

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

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

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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.

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

Cancel common factors.

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

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

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

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

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

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 $12^\\frac{2}{3}$.

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

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

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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.

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

Note: use * for multiplication.

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). Because of this we deal with the different bases separately.

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

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

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

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]}\\]
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]}\\]

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

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

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

Is the following statement true or false?

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

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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True

", "

False

Is the following statement true or false?

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

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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Is the following statement true or false?

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

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.

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

\\[\\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.

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2..6

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

Interpret the powers in expanded form and then write the result in index form.

Suppose you had to simplify $(5^3)^2$. Well cubed means there are three 5s all being multiplied, so we have

\\[(5^3)^2=(5\\times 5\\times 5)^2\\]

and squared means multiplied by itself (in this case the square is acting on the whole bracketed term) so we get

\\[(5^3)^2=(5\\times 5\\times 5)^2=(5\\times 5\\times 5)\\times (5\\times 5\\times 5)\\]

but this is just six 5s all being multiplied together, that is

\\[(5^3)^2=5^6.\\]

In general, we have $\\displaystyle(a^b)^c=a^{bc}$.

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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.

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

Interpret the powers in expanded form and then write the result in index form.

Suppose you had to simplify $(5^3)^2$. Well cubed means there are three 5s all being multiplied, so we have

\\[(5^3)^2=(5\\times 5\\times 5)^2\\]

and squared means multiplied by itself (in this case the square is acting on the whole bracketed term) so we get

\\[(5^3)^2=(5\\times 5\\times 5)^2=(5\\times 5\\times 5)\\times (5\\times 5\\times 5)\\]

but this is just six 5s all being multiplied together, that is

\\[(5^3)^2=5^6.\\]

In general, we have $\\displaystyle(a^b)^c=a^{bc}$.

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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.

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

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

Interpret the powers in expanded form and then write the result in index form.

Suppose you had to simplify $((5^3)^2)^4$. We could start by applying the square

\\[((5^3)^2)^4=(5^3\\times 5^3)^4=(5^6)^4\\]

and then apply the fourth power

\\[(5^6)^4=5^6\\times 5^6\\times 5^6\\times 5^6 = 5^{24}.\\]

Notice the above is the long way of just applying the rule $\\displaystyle(a^b)^c=a^{bc}$ twice:

\\[((5^3)^2)^4=(5^6)^4=5^{24}.\\]

In fact, we could just multiply all the powers together in one go:

\\[((5^3)^2)^4=5^{3\\times 2\\times 4}=5^{24}.\\]

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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.

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

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 $12^\\frac{2}{3}$.

Multiply all the powers together as you have done in previous questions.

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.

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

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

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

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 $12^\\frac{2}{3}$.

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

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

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

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Your answer is longer than necessary.

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

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

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

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

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

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

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 $12^\\frac{2}{3}$.

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

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

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

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Your answer is longer than necessary.

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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.

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

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

By the same reasoning as used in the above questions we have $\\sqrt[n]{a}=a^{\\frac{1}{n}}$.

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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.

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

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

For example, \\[\\sqrt[3]{2}^5=(2^{\\frac{1}{3}})^5=2^{\\frac{5}{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.

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

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

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

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

For example, \\[\\sqrt[3]{2^5}=(2^5)^{\\frac{1}{3}}=2^{\\frac{5}{3}}\\]

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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.

By interpreting the denominator of the fractional power as an n^th root, determine the value of the following:

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

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

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

Convert the denominator of the fractional power to a root.

For example, $27^{\\frac{2}{3}}=\\sqrt[3]{27}^2=3^2=9$.

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

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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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