Questions on manipulating logarithms.
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\n\\[\\log_a(x)+\\log_a(y)=\\log_a(xy)\\text{.}\\]
\nSubstituting in our values for $x$ and $y$ gives
\n\\[\\begin{align}
\\log_a(\\var{x1[1]})+\\log_a(\\var{x1[0]})&=\\log_a(\\var{x1[1]}\\times \\var{x1[0]})\\\\
&=\\log_a(\\var{x1[1]*x1[0]})\\text{.}
\\end{align}\\]
\n
We need to use the rule
\n\\[\\log_a(x)-\\log_a(y)=\\log_a\\left(\\frac{x}{y}\\right)\\text{.}\\]
\nSubstituting in our values for $x$ and $y$ gives
\n\\[\\begin{align}
\\log_a(\\var{x1[4]*y1})-\\log_a(\\var{x1[4]})&=\\log_a(\\var{x1[4]*y1}\\div \\var{x1[4]})\\\\
&=\\log_a(\\var{y1})\\text{.}
\\end{align}\\]
Simplify the expressions to fill in the gaps.
", "variable_groups": [], "parts": [{"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "stepsPenalty": 0, "gaps": [{"correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "type": "numberentry", "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "maxValue": "x1[1]*x1[0]", "showFeedbackIcon": true, "minValue": "x1[1]*x1[0]", "correctAnswerStyle": "plain", "allowFractions": false, "mustBeReduced": false, "scripts": {}, "variableReplacements": [], "marks": "2", "showCorrectAnswer": true}], "showCorrectAnswer": true, "prompt": "$\\log_a(\\var{x1[1]})+ \\log_a(\\var{x1[0]})=\\log_a($ [[0]]$)$
", "steps": [{"scripts": {}, "variableReplacements": [], "type": "information", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "showCorrectAnswer": true, "prompt": "When adding and subtracting logarithms we can simplify the expressions using some logarithm laws. These laws are
\n\\[\\begin{align}
\\log_a(x)+\\log_a(y)&=\\log_a(xy)\\text{,}\\\\
\\log_a(x)-\\log_a(y)&=\\log_a\\left(\\frac{x}{y}\\right)\\text{.}
\\end{align}\\]
$\\log_a(\\var{(x1[4])*y1})-\\log_a(\\var{x1[4]})=\\log_a($ [[0]]$)$
", "steps": [{"scripts": {}, "variableReplacements": [], "type": "information", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "showCorrectAnswer": true, "prompt": "When adding and subtracting logarithms we can simplify the expressions using some logarithm laws. These laws are
\n\\[\\begin{align}
\\log_a(x)+\\log_a(y)&=\\log_a(xy)\\text{,}\\\\
\\log_a(x)-\\log_a(y)&=\\log_a\\left(\\frac{x}{y}\\right)\\text{.}
\\end{align}\\]
Use laws for addition and subtraction of logarithms to simplify a given logarithmic expression to an arbitrary base.
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\n\\[\\log_a(x^y)=y\\log_a(x)\\text{.}\\]
\nThis can also be useful for removing integers from the front of logarithms.
", "advice": "i)
\nWe need to use the rule
\n\\[k\\log_a(x)=\\log_a(x^k)\\text{.}\\]
\nSubsituting in our values for $x$ and $k$ gives
\n\\[\\var{x1[3]}\\log_a(\\var{z1[0]})=\\log_a(\\var{z1[0]^x1[3]})\\text{.}\\]
\nii)
\nWe need to use the rule
\n\\[k\\log_a(x)=\\log_a(x^k)\\text{.}\\]
\nSubsituting in our values for $x$ and $k$ gives
\n\\[\\var{x1[1]}\\log_a(\\var{z1[1]})=\\log_a(\\var{z1[1]^x1[1]})\\text{.}\\]
\ni)
\nThe rule for indices in logarithms also works the other way around,
\n\\[\\log_a(x^k)=k\\log_a(x)\\text{.}\\]
\nWe can use this to rearrange our expression by substituting in values for $x$ and $k$.
\n\\[\\begin{align}
\\log_a(\\var{x1[3]^z1[5]})&=k\\log_a(\\var{x1[3]})\\\\
\\var{x1[3]^z1[5]}&=\\var{x1[3]}^k\\\\
\\var{x1[3]^z1[5]}&=\\var{x1[3]}^\\var{z1[5]}\\\\
k&=\\var{z1[5]}\\\\
\\log_a(\\var{x1[3]^z1[5]})&=\\var{z1[5]}\\log_a(\\var{x1[3]})
\\end{align}\\]
ii)
\nAs with i) we can use the rule
\n\\[\\log_a(x^k)=k\\log_a(x)\\text{.}\\]
\nWe can use this to rearrange our expression by substituting in values for $x$ and $k$.
\n\\[\\begin{align}
\\log_a(\\var{x1[5]^z1[6]})&=k\\log_a(\\var{x1[5]})\\\\
\\var{x1[5]^z1[6]}&=\\var{x1[5]}^k\\\\
\\var{x1[5]^z1[6]}&=\\var{x1[5]}^\\var{z1[6]}\\\\
k&=\\var{z1[6]}\\\\
\\log_a(\\var{x1[5]^z1[6]})&=\\var{z1[6]}\\log_a(\\var{x1[5]})
\\end{align}\\]
i)
\nFrom the structure of this question we can tell that the answer can be written in the form $k\\log_a(\\var{x1[3]})$, meaning all of the values in the expression
\n\\[\\log_a(\\var{x1[3]^z1[2]})+\\log_a(\\var{x1[3]})\\]
\ncan be written in the form $k\\log_a(\\var{x1[3]})$.
\nIf we look at each log individually we can make sure they all take this form.
\n\\[\\begin{align}
\\log_a(\\var{x1[3]^z1[2]})&=k\\log_a(\\var{x1[3]})\\\\
\\var{x1[3]^z1[2]}&=\\var{x1[3]}^k\\\\
\\var{x1[3]^z1[2]}&=\\var{x1[3]}^\\var{z1[2]}\\\\
k&=\\var{z1[2]}\\\\
\\log_a(\\var{x1[3]^z1[2]})&=\\var{z1[2]}\\log_a(\\var{x1[3]})
\\end{align}\\]
We can now write our expression as
\n\\[\\begin{align}
\\log_a(\\var{x1[3]^z1[2]})+\\log_a(\\var{x1[3]})&=\\var{z1[2]}\\log_a(\\var{x1[3]})+\\log_a(\\var{x1[3]})\\\\
&=\\var{z1[2]+1}\\log_a(\\var{x1[3]})\\text{.}
\\end{align}\\]
ii)
\nFrom this question we know our answer is written in the form $k\\log_a(\\var{x1[4]})$, meaning all of the values in the expression
\n\\[\\log_a(\\var{x1[4]^z1[1]})+\\log_a(\\var{x1[4]^z1[0]})\\]
\ncan be written in the form $k\\log_a(\\var{x1[4]})$.
\nIf we look at each log individually we can make sure they all take this form.
\n\\[\\begin{align}
\\log_a(\\var{x1[4]^z1[1]})&=k\\log_a(\\var{x1[4]})\\\\
\\var{x1[4]^z1[1]}&=\\var{x1[4]}^k\\\\
\\var{x1[4]^z1[1]}&=\\var{x1[4]}^\\var{z1[1]}\\\\
k&=\\var{z1[1]}\\\\
\\log_a(\\var{x1[4]^z1[1]})&=\\var{z1[1]}\\log_a(\\var{x1[4]})
\\end{align}\\]
\\[\\begin{align}
\\log_a(\\var{x1[4]^z1[0]})&=k\\log_a(\\var{x1[4]})\\\\
\\var{x1[4]^z1[0]}&=\\var{x1[4]}^k\\\\
\\var{x1[4]^z1[0]}&=\\var{x1[4]}^\\var{z1[0]}\\\\
k&=\\var{z1[0]}\\\\
\\log_a(\\var{x1[4]^z1[0]})&=\\var{z1[0]}\\log_a(\\var{x1[4]})
\\end{align}\\]
We can now write our expression as
\n\\[\\begin{align}
\\log_a(\\var{x1[4]^z1[1]})+\\log_a(\\var{x1[4]^z1[0]})&=\\var{z1[1]}\\log_a(\\var{x1[4]})+\\var{z1[0]}\\log_a(\\var{x1[4]})\\\\
&=\\var{z1[1]+z1[0]}\\log_a(\\var{x1[4]})\\text{.}
\\end{align}\\]
iii)
\nFrom this question we know our answer is written in the form $k\\log_a(\\var{x1[5]})$, meaning all of the values in the expression
\n\\[\\log_a(\\var{x1[5]^z1[1]})+\\log_a(\\var{x1[5]^z1[2]})-\\log_a(\\var{x1[5]^z1[4]})\\]
\ncan be written in the form $k\\log_a(\\var{x1[5]})$.
\nIf we look at each log individually we can make sure they all take this form.
\n\\[\\begin{align}
\\log_a(\\var{x1[5]^z1[1]})&=k\\log_a(\\var{x1[5]})\\\\
\\var{x1[5]^z1[1]}&=\\var{x1[5]}^k\\\\
\\var{x1[5]^z1[1]}&=\\var{x1[5]}^\\var{z1[1]}\\\\
k&=\\var{z1[1]}\\\\
\\log_a(\\var{x1[5]^z1[1]})&=\\var{z1[1]}\\log_a(\\var{x1[5]})
\\end{align}\\]
\\[\\begin{align}
\\log_a(\\var{x1[5]^z1[2]})&=k\\log_a(\\var{x1[5]})\\\\
\\var{x1[5]^z1[2]}&=\\var{x1[5]}^k\\\\
\\var{x1[5]^z1[2]}&=\\var{x1[5]}^\\var{z1[2]}\\\\
k&=\\var{z1[2]}\\\\
\\log_a(\\var{x1[5]^z1[2]})&=\\var{z1[2]}\\log_a(\\var{x1[5]})
\\end{align}\\]
\\[\\begin{align}
\\log_a(\\var{x1[5]^z1[4]})&=k\\log_a(\\var{x1[5]})\\\\
\\var{x1[5]^z1[4]}&=\\var{x1[5]}^k\\\\
\\var{x1[5]^z1[4]}&=\\var{x1[5]}^\\var{z1[4]}\\\\
k&=\\var{z1[4]}\\\\
\\log_a(\\var{x1[5]^z1[4]})&=\\var{z1[4]}\\log_a(\\var{x1[5]})
\\end{align}\\]
We can now write our expression as
\n\\[\\begin{align}
\\log_a(\\var{x1[5]^z1[1]})+\\log_a(\\var{x1[5]^z1[2]})-\\log_a(\\var{x1[5]^z1[4]})&=\\var{z1[1]}\\log_a(\\var{x1[5]})+\\var{z1[0]}\\log_a(\\var{x1[5]})-\\var{z1[4]}\\log_a(\\var{x1[5]})\\\\
&=\\var{z1[1]+z1[2]-z1[4]}\\log_a(\\var{x1[5]})\\text{.}
\\end{align}\\]
Simplify the following expressions.
\ni)
\n$\\var{z1[0]}\\log_a(\\var{x1[3]})=\\log_a($ [[0]]$)$
\nii)
\n$\\var{z1[1]}\\log_a(\\var{x1[1]})=\\log_a($ [[1]]$)$
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\ni)
\n$\\log_a(\\var{x1[3]^z1[5]})=$ [[0]] $\\log_a(\\var{x1[3]})$
\nii)
\n$\\log_a(\\var{x1[5]^z1[6]})=$ [[1]] $\\log_a(\\var{x1[5]})$
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\n$\\log_a(\\var{x1[3]^z1[2]})+\\log_a(\\var{x1[3]})=$ [[0]]$\\log_a(\\var{x1[3]})$
\nii)
\n$\\log_a(\\var{x1[4]^z1[1]})+\\log_a(\\var{x1[4]^z1[0]})=$ [[1]]$\\log_a(\\var{x1[4]})$
\niii)
\n$\\log_a(\\var{x1[5]^z1[1]})+\\log_a(\\var{x1[5]^z1[2]})-\\log_a(\\var{x1[5]^z1[4]})=$ [[2]]$\\log_a(\\var{x1[5]})$
", "marks": 0}], "ungrouped_variables": ["x1", "y1", "z1", "b1", "c", "b4", "b", "b2"], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Use the rule $\\log_a(n^b) = b\\log_a(n)$ to rearrange some expressions.
"}, "preamble": {"css": "", "js": ""}, "functions": {}}, {"name": "Using the Logarithm Equivalence $\\log_ba=c \\Longleftrightarrow a=b^c$", "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": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}], "type": "question", "statement": "Changing the subject of an equation involving logarithms often requires the use of the equivalence
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\text{.}\\]
", "variablesTest": {"condition": "", "maxRuns": 100}, "variables": {"h2": {"group": "part3", "description": "", "templateType": "anything", "name": "h2", "definition": "random(2..4)"}, "f3": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "f3", "definition": "random(3..8)"}, "h1": {"group": "part3", "description": "", "templateType": "anything", "name": "h1", "definition": "random(1..10 except h2)"}, "g2": {"group": "part 2", "description": "", "templateType": "anything", "name": "g2", "definition": "random(2..10except g1)"}, "f2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "f2", "definition": "random(2..10 except f3 f)"}, "f5": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "f5", "definition": "random(2..6 except f1)"}, "f4": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "f4", "definition": "random(5..12 except f2 f)"}, "f1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "f1", "definition": "random(2..5 except f)"}, "g1": {"group": "part 2", "description": "", "templateType": "anything", "name": "g1", "definition": "random(2..10)"}, "f": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "f", "definition": "random(2..10)"}, "g3": {"group": "part 2", "description": "", "templateType": "anything", "name": "g3", "definition": "random(2..10except g1 g2)"}, "g4": {"group": "part 2", "description": "", "templateType": "anything", "name": "g4", "definition": "random(2..10except g1 g2 g3)"}}, "functions": {}, "tags": ["logarithm", "Logarithm", "Logarithm equivalence law", "logarithm laws", "Logs", "logs", "taxonomy"], "variable_groups": [{"name": "part 2", "variables": ["g3", "g2", "g4", "g1"]}, {"name": "part3", "variables": ["h1", "h2"]}], "parts": [{"scripts": {}, "variableReplacements": [], "marks": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"checkingtype": "absdiff", "scripts": {}, "showpreview": true, "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "expectedvariablenames": [], "marks": 1, "variableReplacements": [], "answer": "{f^f1}", "checkvariablenames": false, "type": "jme"}], "showCorrectAnswer": true, "prompt": "Rearrange the equation to find $x$.
\n$\\log_\\var{f}(x)=\\var{f1}$
\n$x=$ [[0]]
", "type": "gapfill"}, {"scripts": {}, "variableReplacements": [], "marks": 0, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"checkingtype": "absdiff", "scripts": {}, "showpreview": true, "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "expectedvariablenames": [], "marks": 1, "variableReplacements": [], "answer": "{g1}^(y+{g2})", "checkvariablenames": false, "type": "jme"}], "showCorrectAnswer": true, "prompt": "Make $x$ the subject of the following equation.
\n$\\log_\\var{g1}(x)=y+\\var{g2}$
\n$x=$ [[0]]
Make $x$ the subject of the equation, leaving your answer in the form $a^{\\frac{1}{b}}$.
\n$\\log_x(y+\\var{h1})=\\var{h2}$
\n$x=$ [[0]]
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", "$a^{\\log_a(x)}$
", "$e^{\\ln(x)}$
", "$\\log_{10}(x)$
", "$\\log_e(x)$
", "$\\ln(e^x)$
"], "showFeedbackIcon": true, "prompt": "Which of the following expressions are equivalent to $x$?
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"}, "preamble": {"css": "", "js": ""}, "advice": "i)
\nWe can rearrange logarithms using indices.
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]
\nUsing this equivalence we can rewrite $\\log_\\var{f}x=\\var{f1}$.
\n\\[\\begin{align}
x&= \\var{f}^\\var{f1} \\\\
&=\\var{f^f1}
\\end{align}\\]
\n
i)
\nWe can use the equivalence to rewrite our equation.
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]
\nWe can write out our values to makes it easier.
\n\\[\\begin{align}
a&=x \\\\
b&=\\var{g1}\\\\
c&=y+\\var{g2}
\\end{align}\\]
Then we can write out our equation in the required form.
\n\\[x=\\var{g1}^{y+\\var{g2}}\\]
\n\n
We can use the same equivalence as in part b).
\n\\[\\log_ba=c \\Longleftrightarrow a=b^c\\]
\nWe have
\n\\begin{align}
a&=y+\\var{h1} \\\\
b&=x\\\\
c&=\\var{h2}\\text{.} \\\\ \\\\
\\log_{x}(y+\\var{h1}) &= \\var{h2} \\\\
\\implies y+\\var{h1} &= x^{\\var{h2}} \\\\
x &= (y+\\var{h1})^{\\frac{1}{\\var{h2}}}
\\end{align}
The two in this list that don't equal $x$ are $\\log_e(x)$ and $\\log_{10}(x)$.
\n\\[\\begin{align}
\\log_e(x)&=\\ln(x)\\\\
\\log_{10}(x)&=\\log(x)\\text{.}
\\end{align}\\]
Apply and combine logarithm laws in a given equation to find the value of $x$.
", "licence": "Creative Commons Attribution 4.0 International"}, "ungrouped_variables": ["x1", "y1", "z1", "b1", "c", "b4", "b", "b2"], "type": "question", "rulesets": {}, "advice": "We can use the logarithm law
\n\\[k\\log_a(x)=\\log_a(x^k)\\text{,}\\]
\nto also give a more specific rule
\n\\[\\begin{align}
\\log_a\\left(\\frac{1}{x}\\right)&=\\log_a(x^{-1})\\\\
&=-\\log_a(x)\\text{.}
\\end{align}\\]
This means we can write our expression as
\n\\[\\log_\\var{b1}(x-\\var{b2})+\\log_\\var{b1}({x})=\\var{b4}\\text{.}\\]
\nThen using the rule
\n\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\text{,}\\]
\nwe can write our equation as
\n\\[\\begin{align}
\\log_\\var{b1}(x(x-\\var{b2}))&=\\var{b4}\\\\
\\log_\\var{b1}(x^2-\\var{b2}x)&=\\var{b4}\\text{.}\\\\
\\end{align}\\]
We then rely on the definition of $\\log_a$
\n\\[b=a^c \\Longleftrightarrow \\log_{a}b=c\\]
\nto write our equation as
\n\\[\\begin{align}
x^2-\\var{b2}x&=\\var{b1}^\\var{b4}\\\\
&=\\var{b1^b4}\\text{.}
\\end{align}\\]
We can then write out our equation and solve either by factorising or using the quadratic formula;
\n\\[\\begin{align}
x^2-\\var{b2}x-\\var{b1^{b4}}&=0\\\\
(x+2)(x-\\var{b})&=0\\text{.}
\\end{align}\\]
As logarithms can only be applied to positive numbers, the only possible value for $x$ is $\\var{b}$.
\n$\\ln(x)$ is a shorthand for $\\log_e(x)$, so we can apply the same laws of logarithms here.
\nTherefore applying the rule
\n\\[k\\log_a(x)=\\log_a(x^k)\\]
\nwe can write our equation as
\n\\[\\ln(x^\\var{p})+\\ln(\\var{q})=\\var{m}\\text{.}\\]
\nThen using the rule
\n\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\]
\nwe can write our equation as
\n\\[\\ln(\\var{q}x^\\var{p})=\\var{m}\\text{.}\\]
\nAs $\\ln=\\log_e$ we can use
\n\\[a=b^c \\Longleftrightarrow \\log_ba=c\\]
\nto write our equation as
\n\\[\\var{q}x^\\var{p}=e^\\var{m}\\text{.}\\]
\nWe then just need to rearrange our equation
\n\\[\\begin{align}
\\var{q}x^\\var{p}&=e^\\var{m}\\\\[0.5em]
x^\\var{p}&=\\frac{e^\\var{m}}{\\var{q}}\\\\[0.5em]
x&=\\frac{e^{\\var{m}/\\var{p}}}{\\var{q^(1/{p})}}
\\end{align}\\]
Solve for $x$.
\n$\\log_\\var{b1}(x-\\var{b2})-\\log_\\var{b1}\\left(\\displaystyle\\frac{1}{x}\\right)=\\var{b4}$
\n$x=$ [[0]]
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