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Questions on powers, the laws of indices, and exponential growth.
", "licence": "Creative Commons Attribution 4.0 International"}, "name": "Nick's copy of Powers and indices", "timing": {"timedwarning": {"message": "", "action": "none"}, "allowPause": true, "timeout": {"message": "", "action": "none"}}, "question_groups": [{"pickingStrategy": "all-ordered", "name": "Group", "pickQuestions": 1, "questions": [{"name": "Calculate powers of ten", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "advice": "When $n$ is positive, we multiply $10$ by itself $n$ times,
\n\\[\\text{e.g. } 10^3 = 10 \\times 10 \\times 10 = 1000 \\text{ .}\\]
\nWhen $n$ is negative, we can think of $10^{-n}$ as $\\frac{1}{10^{n}}$,
\n\\[\\text{e.g. } 10^{-3} = \\frac{1}{10^3} = \\frac{1}{1000} = 0.001\\text{ .}\\]
\nWhen $n = 0$:
\n\\[10^{0} = 1 \\text{ .}\\]
\nGenerally, we can think of $10^n$ as a number in standard form $1 \\times 10^n$. Then $n$ always tells us the number of decimal places to move the decimal point in $1.0$, for example
\n\\[10^{-3} = 1.0 \\times 10^{-3} \\text{ and since } n = - 3 \\text{, we go } 3 \\text{ places back as follows: } 1.0 ⇒ 0.1 ⇒ 0.01 ⇒ 0.001 \\text{ .}\\]
\nA complete table of powers of ten for $n$ from $-6$ to $6$ is:
\n$n$ | \n$10^n$ | \n
---|---|
$-6$ | \n$0.000001$ | \n
$-5$ | \n$0.00001$ | \n
$-4$ | \n$0.0001$ | \n
$-3$ | \n$0.001$ | \n
$-2$ | \n$0.01$ | \n
$-1$ | \n$0.1$ | \n
$0$ | \n$1$ | \n
$1$ | \n$10$ | \n
$2$ | \n$100$ | \n
$3$ | \n$1000$ | \n
$4$ | \n$10000$ | \n
$5$ | \n$100000$ | \n
$6$ | \n$1000000$ | \n
Powers of ten can be useful while working with standard index numbers. Fill in the following table of powers of ten:
", "variables": {"n": {"name": "n", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(2..5)"}}, "tags": ["powers", "powers of 10", "standard index numbers", "taxonomy"], "ungrouped_variables": ["n"], "functions": {}, "preamble": {"js": "", "css": ""}, "type": "question", "variable_groups": [], "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": "1000"}, "metadata": {"description": "Fill in a table of powers of 10.
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---|---|
[[0]] | \n$\\var{10^(-n-1)}$ | \n
$\\var{-n+1}$ | \n[[1]] | \n
$0$ | \n$1$ | \n
$1$ | \n$10$ | \n
$\\var{n}$ | \n[[2]] | \n
[[3]] | \n$\\var{10^(n+2)}$ | \n
Sorted list of integers from 1 to 12.
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", "name": "uy", "group": "Ungrouped variables"}, "y": {"templateType": "anything", "definition": "random(3..12 except x)", "description": "Answer to part c).
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\n$\\var{x[0]}^2 =$ [[0]]
\n$\\var{x[1]}^2 =$ [[1]]
\n$\\var{x[2]}^2 =$ [[2]]
\n$\\var{x[3]}^2 =$ [[3]]
\n$\\var{x[4]}^2 =$ [[4]]
\n"}, {"showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[0]*x[0]*x[0]", "maxValue": "x[0]*x[0]*x[0]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[1]*x[1]*x[1]", "maxValue": "x[1]*x[1]*x[1]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[2]*x[2]*x[2]", "maxValue": "x[2]*x[2]*x[2]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[3]*x[3]*x[3]", "maxValue": "x[3]*x[3]*x[3]", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "x[4]*x[4]*x[4]", "maxValue": "x[4]*x[4]*x[4]", "marks": 1, "variableReplacements": []}], "prompt": "Find the following:
\n$\\var{x[0]}^3 =$ [[0]]
\n$\\var{x[1]}^3 =$ [[1]]
\n$\\var{x[2]}^3 =$ [[2]]
\n$\\var{x[3]}^3 =$ [[3]]
\n$\\var{x[4]}^3 =$ [[4]]
"}, {"showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "y^2", "maxValue": "y^2", "marks": 1, "variableReplacements": []}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "scripts": {}, "minValue": "y", "maxValue": "y", "marks": "1", "variableReplacements": []}], "prompt": "Find a square number $y^2$ between $\\var{ly}$ and $\\var{uy}$ and its integer root $y$.
\n$y^2 = $ [[0]]
\n$y = $ [[1]]
"}], "advice": "Squared integers are called square numbers. It may be useful to remember the first few square numbers to be able to use them without a calculator.
\nHere:
\n\\[ \\begin{align} \\var{x[0]}^2 &= \\var{x[0]} \\times \\var{x[0]} \\\\&= \\var{x[0]^2} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[1]}^2 &= \\var{x[1]} \\times \\var{x[1]} \\\\&= \\var{x[1]^2} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[2]}^2 &= \\var{x[2]} \\times \\var{x[2]} \\\\&= \\var{x[2]^2} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[3]}^2 &= \\var{x[3]} \\times \\var{x[3]} \\\\&= \\var{x[3]^2} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[4]}^2 &= \\var{x[4]} \\times \\var{x[4]} \\\\&= \\var{x[4]^2} \\text{.}\\end{align}\\]
\n\nCubed integers are called cubed numbers. To obtain these, we would typically always use a calculator.
\nWe can either cube the number $x$:
\n\\[ \\begin{align} \\var{x[0]}^3 &= \\var{x[0]} \\times \\var{x[0]} \\times \\var{x[0]} \\\\&= \\var{x[0]^3} \\text{,} \\end{align}\\]
\nor we can multiply the square number $(x_n)^2$ from part a) by the appropriate $x_n$:
\n\\[ \\begin{align} \\var{x[0]}^3 &= \\var{x[0]}^2 \\times \\var{x[0]} \\\\&= \\var{x[0]^2} \\times \\var{x[0]}\\\\&= \\var{x[0]^3} \\text{.} \\end{align}\\]
\n\\[ \\begin{align} \\var{x[1]}^3 &= \\var{x[1]} \\times \\var{x[1]} \\times \\var{x[1]} \\\\ &= \\var{x[1]^3} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[2]}^3 &= \\var{x[2]} \\times \\var{x[2]} \\times \\var{x[2]} \\\\ &= \\var{x[2]^3} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[3]}^3 &= \\var{x[3]} \\times \\var{x[3]} \\times \\var{x[3]} \\\\ &= \\var{x[3]^3} \\text{.}\\end{align}\\]
\n\\[ \\begin{align} \\var{x[3]}^4 &= \\var{x[4]} \\times \\var{x[4]} \\times \\var{x[4]} \\\\ &= \\var{x[4]^3} \\text{.}\\end{align}\\]
\n\nHere is a table of square numbers for integers from $1$ to $15$:
\n$y$ | \n$1$ | \n$2$ | \n$3$ | \n$4$ | \n$5$ | \n$6$ | \n$7$ | \n$8$ | \n$9$ | \n$10$ | \n$11$ | \n$12$ | \n$13$ | \n$14$ | \n$15$ | \n
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
$y^2$ | \n$1$ | \n$4$ | \n$9$ | \n$16$ | \n$25$ | \n$36$ | \n$49$ | \n$64$ | \n$81$ | \n$100$ | \n$121$ | \n$144$ | \n$169$ | \n$196$ | \n$225$ | \n
The only square number between $\\var{ly}$ and $\\var{uy}$ is $\\var{y2}$.
\nTo calculate $y$ we must calculate the square root of $y^2$,
\n\\[ \\sqrt{\\var{y2}} = \\var{y} \\text{.}\\]
\nThis is our integer $y$.
\n\n", "tags": ["cube", "indices", "multiplication", "powers", "roots", "square", "taxonomy"], "preamble": {"js": "", "css": ""}, "rulesets": {}, "functions": {}, "ungrouped_variables": ["x1", "x2", "x3", "x4", "x5", "x", "y2", "y", "ly", "uy"], "statement": "Try the following questions on square and cube numbers.
", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Find the squares, and cubes, of some numbers.
\nFinally, find a square number between two given limits.
"}, "variablesTest": {"condition": "", "maxRuns": "1000"}}, {"name": "Always, sometimes or never: square and cube numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "type": "question", "statement": "For each of the following statements, select one option from \"Always\", \"Sometimes\" or \"Never\".
\nSelect:
\nThis variable was created solely for the purpose of being able to publish this question.
", "templateType": "anything", "definition": "1"}}, "functions": {}, "tags": ["always", "cubic numbers", "never", "sometimes", "square numbers", "taxonomy"], "variable_groups": [], "parts": [{"layout": {"expression": "", "type": "all"}, "scripts": {}, "minMarks": 0, "type": "m_n_x", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "displayType": "radiogroup", "choices": ["i) $x^3$ is greater than $x^2$
", "ii) $x^2$ is greater than $x$
", "iii) If $x$ is negative, $x^2$ is negative
", "iv) If $x$ is negative, $x^3$ is negative
", "v) $x^2 = x$
", "vi) $x^2 = - x$
", "vii) $(x+1)^2 \\gt x$
", "viii) $(x+1)^3 \\gt x$
", "ix) $x^3 \\times x = x^2 \\times x^2$
", "x) $x^2$ has the opposite sign to $x$
", "xi) $x^3$ has the opposite sign to $x$
"], "showFeedbackIcon": true, "answers": ["Always
", "Sometimes
", "Never
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"}, "preamble": {"css": "", "js": ""}, "advice": "i)
\nSuppose $x$ is negative, for example $x = -5$.
\n\\[ \\begin{align} x^2 &= (-5)^2 \\\\&= 25 \\\\ x^3 &= (-5)^3 \\\\&= -125 \\\\x^3 &\\lt x^2\\text{.} \\end{align} \\]
\nNow let $x$ be positive, for example $x = 5$.
\n\\[ \\begin{align} x^2 &= 5^2 \\\\&= 25 \\\\ x^3 &= 5^3 \\\\&= 125 \\\\x^3 &\\gt x^2\\text{.}\\end{align} \\]
\nTherefore, $x^3$ is sometimes greater than $x^2$.
\n\n
\n
ii)
\nThis is true for either $x \\gt 1$ or $x \\lt 0$ but false for $0 \\leq x \\leq 1$. For example, let $x=0.5$. Then
\n\\[ \\begin{align} \\text{When } x &= 0.5\\text{,} \\\\x^2 &= 0.25\\text{, so} \\\\ x^2 &\\lt x \\text{.} \\end{align} \\]
\nTherefore, $x^2$ is sometimes greater than $x$.
\n\n
\n
iii)
\nMultiplying two negative numbers gives a positive answer and multiplying two postive numbers gives a positive answer. Therefore, $x^2$ is never negative.
\n\n
\n
iv)
\nMultiplying a negative number by itself an odd number of times always gives a negative answer. For example, let $x = -1$. Then
\n\\[\\begin{align} x^3 &= (-1)^3 \\\\&= -1\\times-1\\times-1 \\\\&=1\\times-1 \\\\&= - 1 \\text{.} \\end{align}\\]
\nTherefore, if $x$ is negative, $x^3$ is always negative.
\n\n
\n
v)
\nThis is true when $x = 1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $x$.
\n\n
\n
vi)
\nThis is true when $x = -1$ but false for all other values of $x$. Therefore, $x^2$ sometimes equals $-x$.
\n\n
\n
vii)
\nWhen $x$ is positive, for example $x = 5$:
\n\\[ \\begin{align} (x+1)^2 &= (5 + 1)^2 \\\\&= 6^2 \\\\&= 36 \\gt x = 5 \\end{align} \\]
\nWhen $x = 0$:
\n\\[ \\begin{align} (x+1)^2 &= (0 + 1)^2 \\\\&= 1^2 \\\\&= 1 \\gt x = 0 \\end{align} \\]
\nWhen $x$ is negative, such as $x = -4$:
\n\\[ \\begin{align} (x+1)^2 &= (-4 + 1)^2 \\\\&= (-3)^2 \\\\&= 9 \\gt x = -4 \\end{align} \\]
\nTo see the behaviour of $(x+1)^2$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:
\n$x$ | \n$-3$ | \n$-2$ | \n$-1$ | \n$-0.5$ | \n$0$ | \n$0.5$ | \n$1$ | \n$2$ | \n\n $3$ \n | \n
---|---|---|---|---|---|---|---|---|---|
$(x+1)^2$ | \n$4$ | \n$1$ | \n$0$ | \n$0.25$ | \n$1$ | \n$2.25$ | \n$4$ | \n$9$ | \n$16$ | \n
Therefore, $(x+1)^2$ is always greater than $x$.
\n\n\n
viii)
\nWhen $x$ is positive, for example $x = 5$:
\n\\[ \\begin{align} (x+1)^3 &= (5 + 1)^3 \\\\&= 6^3 \\\\&= 216 \\gt x = 5 \\end{align} \\]
\nWhen $x = 0$:
\n\\[ \\begin{align} (x+1)^3 &= (0 + 1)^3 \\\\&= 1^3 \\\\&= 1 \\gt x = 0 \\end{align} \\]
\nWhen $x$ is negative, such as $x = -4$:
\n\\[ \\begin{align} (x+1)^2 &= (-4 + 1)^3 \\\\&= (-3)^3 \\\\&= -27 \\lt x = -4 \\end{align} \\]
\nTo see the behaviour of $(x+1)^3$ a bit more clearly, we make a table for values $-3 \\leq x \\leq 3$:
\n$x$ | \n$-3$ | \n$-2$ | \n$-1$ | \n$-0.5$ | \n$0$ | \n$0.5$ | \n$1$ | \n$2$ | \n\n $3$ \n | \n
---|---|---|---|---|---|---|---|---|---|
$(x+1)^3$ | \n$-8$ | \n$-1$ | \n$0$ | \n$0.125$ | \n$1$ | \n$3.375$ | \n$8$ | \n$27$ | \n$64$ | \n
Therefore, $(x+1)^3$ is sometimes greater than $x$.
\n\n\nix)
\nWe can write
\n\\[x^3\\times x = x \\times x \\times x \\times x = x^2 \\times x^2\\text{.}\\]
\nTherefore, $x^3 \\times x$ always equals $x^2 \\times x^2$.
\n\n\nx)
\nSince $x^2$ is always positive, $x^2$ only has the opposite sign to $x$ when $x$ is negative.
\nTherefore, $x^2$ sometimes has the opposite sign to $x$.
\n\n\nxi)
\nAs seen in part iv), multiplying a negative number by itself an odd number of times always gives a negative answer. It is also true that multiplying a positive number by itself an odd number of times will give a positive answer. So, $x^3$ always has the same sign as $x$, since we have an odd power.
\nTherefore, $x^3$ never has the opposite sign to $x$.
\n\n"}, {"name": "Laws of Indices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "tags": ["indices", "laws of indices", "powers", "taxonomy"], "metadata": {"description": "
This question aims to test understanding and ability to use the laws of indices.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Using the laws of indices, simplify each expression down to its simplest form. Recall that $a^{0} = 1$ for any number $a$.
", "advice": "Here we are using the rule of indices: $a^m \\times a^n = a^{m+n}$.
\nUsing this rule,
\n\\[
\\begin{align}
a^\\var{x} \\times a^\\var{y}\\ &= a^\\simplify[all, !collectNumbers]{{x}+{y}}\\\\
&= a^\\var{x+y}.
\\end{align}
\\]
We are asked to find $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$.
\nNotice there is a constant in front of each of the terms.
\nTo do this, write the product out explicitly, as
\n\\[\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} = \\var{c} \\times \\var{d} \\times a^\\var{p} \\times a^\\var{q}.\\]
\nWe know that $\\var{c} \\times \\var{d} = \\var{c*d}$, and using the rule of indices: $a^\\var{p} \\times a^\\var{q} = a^\\var{p+q}$.
\nTherefore:
\n\\begin{align}
\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}&= \\var{c*d} \\times a^\\var{p+q} \\\\
&= \\simplify{{c*d}*a^{p+q}}.
\\end{align}
Here we are using: $a^m \\div a^n = a^{m-n}$.
\nWe are asked to simplify the expression, $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$.
\nTo do this, we just have to use the previously mentioned rule of indices. We write this out explicity as
\n\\[\\simplify{{b}*a^{x}/({g}*a^{y})} = \\simplify{{b}/{g}} \\times \\simplify{a^{x}/(a^{y})}.\\]
\nUsing rules of indices,
\n\\begin{align} \\frac{a^\\var{x}}{a^\\var{y}} &= a^\\var{x} \\div a^\\var{y}\\\\
&= a^\\simplify[all, !collectNumbers]{{x}-{y}}\\\\
&= a^\\var{x-y}.
\\end{align}
Therefore,
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\simplify{a^{{x}-{y}}}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
Alternatively,
\nUsing the rule of indices: $a^{-m} = \\displaystyle\\frac{1}{a^{m}}$, we can rewrite the question as:
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\frac{a^\\var{x}}{a^\\var{y}}\\\\
&= \\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}}.
\\end{align}
And then using the rule: $a^m \\times a^n = a^{m+n}$, this becomes:
\n\\begin{align}
\\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}} &= \\simplify{{b}/{g}} \\times a^\\simplify[all,!collectNumbers]{{x}+(-{y})}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
The question asks us to simplify $(\\simplify{{c}*a^{p}})^{\\var{q}}$.
\nTo do this we use the rules:
\n\\[(a^{m})^{n} = a^{mn},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can then expand the equation as
\n\\[(\\simplify{{c}*a^{p}})^{\\var{q}}= \\var{c}^{\\var{q}} \\times (a^{\\var{p}})^{\\var{q}}.\\]
\nThen using the rule of indices mentioned previously,
\n\\[
\\begin{align}
(\\simplify{{c}*a^{p}})^{\\var{q}}&= \\simplify{{c}^{q}} \\times a^\\var{p*q}\\\\
&= \\simplify{{c}^{q}*a^{p*q}}.
\\end{align}
\\]
The question asks us to simplify $\\sqrt[\\var{d}]{\\var{x}^\\var{d}a}$.
\nTo do this we use the rules:
\n\\[a^\\frac{1}{m} = \\sqrt[m]{a},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{d}]{a} &= (\\simplify{a})^\\frac{1}{\\var{d}}\\\\
&= a^\\frac{1}{\\var{d}}.
\\end{align}
\\]
The question requires us to simplify $\\sqrt[\\var{c}]{a^\\var{q}}$.
\nHere, we use the rule of indices: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$, allowing us to expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{c}]{\\simplify{a^{q}}} &= \\simplify[fractionnumbers,all]{(a^{q})^{{1}/{{c}}}}\\\\
&= \\simplify[fractionnumbers,all]{a^{{q}/{c}}}.
\\end{align}
\\]
Used in part c
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\n$a^{-m} = \\displaystyle\\frac{1}{a^m}$.
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", "nameToCompare": ""}, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}], "type": "question"}, {"name": "Working with standard index form ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "type": "question", "statement": "Working with numbers that are very large or very small can be tricky.
\nStandard form allows us to simplify these numbers, using powers of 10.
\n\n
Write the following in standard index form (for example, for $2.01\\times 10^5$ we would write 2.01*10^5
in the gap).
$\\var{A[2]*10^2} = $ [[0]]
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", "type": "gapfill"}], "ungrouped_variables": ["A3dp", "A5dp", "small5", "A", "ran", "B", "int"], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Convert a variety of numbers from decimal to standard index form.
"}, "preamble": {"css": "", "js": ""}, "advice": "Converting from decimal to a standard form, we are looking for $A \\times 10^n$.
\nWe need make the first number ($A$) between 1 and 10, so we put the decimal place after the first non-zero digit.
\n\n
a)
\nIn $\\var{A[2]*10^2}$, the first non-zero digit is $\\var{siground(A[2] - 0.5, 1)}$ so we get $A = \\var{A[2]}$.
\nIf we moved the decimal place in $\\var{A[2]}$ so it matches our original number $\\var{A[2]*10^2}$, we would go 2 places to the right, so $n = 2$.
\n\n
b)
\nIn $\\var{A3dp*10^(-1)}$, the first non-zero digit is $\\var{siground(A3dp - 0.5, 1)}$ so we get $A = \\var{A3dp}$.
\nIf we moved the decimal place in $\\var{A3dp}$ so it matches our original number $\\var{A3dp*10^(-1)}$, we would go 1 place to the left, so $n = -1$.
\n\n
c)
\nIn $\\var{precround(A5dp*10^7,0)}$ the first non-zero digit is $\\var{siground(A5dp - 0.5, 1)}$ so we get $A = \\var{A5dp}$.
\nIf we moved the decimal place in $\\var{A5dp}$ so it matches our original number $\\var{precround(A5dp*10^7,0)}$, we would go 7 places to the right, so $n = 7$.
\n\n
In $\\var{small5}$ the first non-zero digit is {siground({{small5}*10^5} - 0.5, 1)} so we get $A = \\var{small5*10^5}$.
\nIf we moved the decimal place in $\\var{small5*10^5}$ so it matches our original number $\\var{small5}$, we would go 5 places to the left, so $n = -5$.
\n"}, {"name": "Simple interest", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "metadata": {"description": "Calculate the interest accrued in a savings account, given the initial balance and annual interest rate.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Suppose you put £{money} into a savings account exactly {years} years ago and you haven't touched the money since. The simple interest rate on the account is {perc2}% per year.
", "variables": {"perc2": {"name": "perc2", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(0..2.5 #0.05)"}, "money": {"name": "money", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(100..5000 #100)"}, "years": {"name": "years", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(2..6)"}}, "tags": ["interests", "percentages", "taxonomy"], "ungrouped_variables": ["perc2", "money", "years"], "functions": {}, "preamble": {"js": "", "css": ""}, "advice": "This is simple interest, which means the amount added each year is a percentage of the original amount. The amount we add is fixed for all {years} years.
\nFirst, we work out the amount of interest for one year:
\n\\begin{align}
\\var{perc2} \\text{% of } \\var{money} &= \\frac{\\var{perc2}}{100} \\times \\var{money} \\\\
&= \\var{perc2/100} \\times \\var{money} \\\\
&= £\\var{dpformat(perc2/100*money,2)} \\text{.}
\\end{align}
The money has been in the account for {years} years, so we multiply $£\\var{dpformat(perc2/100*money,2)}$ by $\\var{years}$.
\n\\[ £\\var{dpformat(perc2/100*money,2)} \\times \\var{years} = £\\var{dpformat(perc2/100*money*years,2)} \\text{.} \\]
\nAdding this to the original balance:
\n\\[ £\\var{money} + £\\var{dpformat(perc2/100*money*years,2)} = £\\var{dpformat(perc2/100*money*years + money,2)} \\text{.} \\]
\nThis is the amount we would get if we withdrew the whole savings balance today.
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\n£ [[0]]
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\n£ [[0]]
", "steps": [{"scripts": {}, "variableReplacements": [], "type": "information", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "showFeedbackIcon": true, "prompt": "The original price of the phone is $£\\var{price}$ and we are told that the price decreases by $\\var{perc}$% every month.
\n", "marks": 0}, {"correctAnswerFraction": false, "scripts": {}, "maxValue": "1-{perc}/100", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "prompt": "What is the decimal multiplier for the decrease in the smartphones each month?
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", "precisionPartialCredit": 0, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "prompt": "Multiply your answer above by the decimal multiplier to obtain the price after 2 months.
\nNote that this is the same as multiplying the original price by $d^2$, where $d$ is the decimal multiplier.
", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "{price}*(1-{perc}/100)^2", "variableReplacements": [], "marks": "0.5", "strictPrecision": false, "showCorrectAnswer": true, "correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "allowFractions": false, "precision": "2", "maxValue": "{price}*(1-{perc}/100)^2", "precisionType": "dp", "correctAnswerStyle": "plain", "showPrecisionHint": true}], "marks": 0}, {"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "scripts": {}, "maxValue": "n-5", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "minValue": "n-5", "variableReplacements": [], "marks": "2", "showCorrectAnswer": true, "type": "numberentry"}], "showFeedbackIcon": true, "prompt": "{person['name']} has $£\\var{threshold}$ to spend on a smartphone. After how many more full months will {person['pronouns']['they']} be able to afford the smartphone?
\n[[0]] months
", "marks": 0}], "ungrouped_variables": [], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Given the original price of a smartphone and the rate at which it decreases, calculate its price after a given number of months. In the second part, calculate the time remaining until the price goes below a certain point.
"}, "preamble": {"css": "", "js": ""}, "functions": {}, "advice": "We can use the multiplier method to calculate the new price. If the price decreases by {perc}%, this its value is {100-perc}% of the original value after 1 month. The decimal multiplier for {100-perc}% is
\n\\[\\frac{\\var{100-perc}}{100} = \\var{(100-perc)/100} \\text{.}\\]
\nEach month our smartphone's value can be found by multiplying the previous month's value by the decimal multiplier. For example, after the first month, the value is
\n\\[ \\var{(100-perc)/100} \\times\\mathrm{£}\\var{price} = \\mathrm{£}\\var{dpformat(price*(100-perc)/100,2)}\\text{.} \\]
\nTo calculate the price after 5 months, we multiply the original price of the smartphone by our multiplier 5 times:
\n\\[ \\begin{align} \\text{Final worth} &= \\var{price} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\times \\var{(100-perc)/100} \\\\&= \\var{price} \\times \\var{(100-perc)/100}^{5} \\\\&= £\\var{precround(price*((100-perc)/100)^5, 2)} {.} \\end{align}\\]
\nFrom part a), the value after 5 months is £$\\var{precround(price*((100-perc)/100)^5, 2)}$. Continuing to multiply the price by the decimal multiplier,
\n\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100} = £\\var{precround(precround(price*((100-perc)/100)^5, 2)*(100-perc)/100, 2)}\\]
\n\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100}^2 = £\\var{precround(precround(price*((100-perc)/100)^6, 2)*(100-perc)/100, 2)}\\]
\n\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100}^3 = £\\var{precround(precround(price*((100-perc)/100)^7, 2)*(100-perc)/100, 2)}\\]
\n\\[£\\var{precround(price*((100-perc)/100)^5, 2)} \\times \\var{(100-perc)/100}^4 = £\\var{precround(precround(price*((100-perc)/100)^8, 2)*(100-perc)/100, 2)}\\]
\n\nThe smartphone's value will be below $£\\var{threshold}$ after {n-5} more months ({n} months in total since its release).
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