// 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,

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

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

#### a)

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i)

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We need to use the rule

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

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Subsituting in our values for $x$ and $k$ gives

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

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ii)

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We need to use the rule

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

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Subsituting in our values for $x$ and $k$ gives

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

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#### b)

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i)

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

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

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We can use this to rearrange our expression by substituting in values for $x$ and $k$.

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\\\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}\

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ii)

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As with i) we can use the rule

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

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We can use this to rearrange our expression by substituting in values for $x$ and $k$.

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\\\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}\

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#### c)

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i)

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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

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

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

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If we look at each log individually we can make sure they all take this form.

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\\\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}\

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We can now write our expression as

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\\\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}\

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ii)

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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

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

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

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If we look at each log individually we can make sure they all take this form.

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\\\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}\

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\\\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}\

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We can now write our expression as

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\\\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}\

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iii)

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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

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\$\\log_a(\\var{x1[5]^z1[1]})+\\log_a(\\var{x1[5]^z1[2]})-\\log_a(\\var{x1[5]^z1[4]})\$

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

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If we look at each log individually we can make sure they all take this form.

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\\\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}\

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\\\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}\

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\\\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}\

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We can now write our expression as

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\\\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}\

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Simplify the following expressions.

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i)

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

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ii)

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

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Simplify the following expressions.

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i)

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

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ii)

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

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i)

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

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ii)

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

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iii)

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$\\log_a(\\var{x1[5]^z1[1]})+\\log_a(\\var{x1[5]^z1[2]})-\\log_a(\\var{x1[5]^z1[4]})=$ [[2]]$\\log_a(\\var{x1[5]})$

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Use the rule $\\log_a(n^b) = b\\log_a(n)$ to rearrange some expressions.

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