// Numbas version: exam_results_page_options {"name": "Rearranging Logarithms involving Indices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Rearranging Logarithms involving Indices", "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question", "statement": "

When logarithms involve indices we can rearrange them using the rule,

\n

\\[\\log_a(x^y)=y\\log_a(x)\\text{.}\\]

\n

This can also be useful for removing integers from the front of logarithms.

", "advice": "

a)

\n

i)

\n

We need to use the rule

\n

\\[k\\log_a(x)=\\log_a(x^k)\\text{.}\\]

\n

Subsituting 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{.}\\]

\n

ii)

\n

We need to use the rule

\n

\\[k\\log_a(x)=\\log_a(x^k)\\text{.}\\]

\n

Subsituting 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{.}\\]

\n

b)

\n

i)

\n

The rule for indices in logarithms also works the other way around,

\n

\\[\\log_a(x^k)=k\\log_a(x)\\text{.}\\]

\n

We 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}\\]

\n

ii)

\n

As with i) we can use the rule

\n

\\[\\log_a(x^k)=k\\log_a(x)\\text{.}\\]

\n

We 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}\\]

\n

c)

\n

i)

\n

From 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]})\\]

\n

can be written in the form $k\\log_a(\\var{x1[3]})$.

\n

If 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}\\]

\n

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}\\]

\n

ii)

\n

From 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]})\\]

\n

can be written in the form $k\\log_a(\\var{x1[4]})$.

\n

If 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}\\]

\n

\\[\\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}\\]

\n

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}\\]

\n

iii)

\n

From 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]})\\]

\n

can be written in the form $k\\log_a(\\var{x1[5]})$.

\n

If 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}\\]

\n

\\[\\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}\\]

\n

\\[\\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}\\]

\n

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}\\]

", "variables": {"v": {"description": "", "group": "part c", "templateType": "anything", "name": "v", "definition": "random(2..10)"}, "b4": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "b4", "definition": "random(2..4 except b1)"}, "c": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "c", "definition": "b1^b4"}, "m": {"description": "", "group": "part c", "templateType": "anything", "name": "m", "definition": "random(2..20)"}, "p": {"description": "", "group": "part c", "templateType": "anything", "name": "p", "definition": "random(3..6)"}, "x1": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "x1", "definition": "repeat(random(2..5),8)"}, "q": {"description": "", "group": "part c", "templateType": "anything", "name": "q", "definition": "v^p"}, "y1": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "y1", "definition": "random(2..6)"}, "b1": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "b1", "definition": "random(2..8 #2)"}, "b": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "b", "definition": "c/2"}, "b2": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "b2", "definition": "b-2"}, "z1": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "name": "z1", "definition": "repeat(random(2..4),10)"}}, "extensions": [], "tags": ["logarithm", "Logarithm", "Logs", "logs", "taxonomy"], "variable_groups": [{"name": "part c", "variables": ["p", "v", "q", "m"]}], "parts": [{"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{x1[3]}^{z1[0]}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}, {"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{x1[1]}^{z1[1]}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}], "showCorrectAnswer": true, "prompt": "

Simplify the following expressions. 

\n

i)

\n

$\\var{z1[0]}\\log_a(\\var{x1[3]})=\\log_a($ [[0]]$)$

\n

ii)

\n

$\\var{z1[1]}\\log_a(\\var{x1[1]})=\\log_a($ [[1]]$)$

", "marks": 0}, {"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{z1[5]}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}, {"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{z1[6]}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}], "showCorrectAnswer": true, "prompt": "

Simplify the following expressions. 

\n

i)

\n

$\\log_a(\\var{x1[3]^z1[5]})=$ [[0]] $\\log_a(\\var{x1[3]})$

\n

ii)

\n

$\\log_a(\\var{x1[5]^z1[6]})=$ [[1]] $\\log_a(\\var{x1[5]})$

", "marks": 0}, {"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "gaps": [{"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{z1[2]+1}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}, {"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{z1[1]+z1[0]}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}, {"checkingtype": "absdiff", "scripts": {}, "type": "jme", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "showCorrectAnswer": true, "showFeedbackIcon": true, "expectedvariablenames": [], "vsetrange": [0, 1], "checkingaccuracy": 0.001, "answer": "{z1[1]+z1[2]-z1[4]}", "marks": 1, "variableReplacements": [], "checkvariablenames": false, "showpreview": true}], "showCorrectAnswer": true, "prompt": "

i)

\n

$\\log_a(\\var{x1[3]^z1[2]})+\\log_a(\\var{x1[3]})=$ [[0]]$\\log_a(\\var{x1[3]})$

\n

ii)

\n

$\\log_a(\\var{x1[4]^z1[1]})+\\log_a(\\var{x1[4]^z1[0]})=$ [[1]]$\\log_a(\\var{x1[4]})$

\n

iii)

\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": {}, "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/"}]}]}], "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/"}]}