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Questions on manipulating logarithms.

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#### a)

\n

We need to use the rule

\n

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

\n

Substituting 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

\n

#### b)

\n

We need to use the rule

\n

\$\\log_a(x)-\\log_a(y)=\\log_a\\left(\\frac{x}{y}\\right)\\text{.}\$

\n

Substituting 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}\

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Simplify the expressions to fill in the gaps.

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

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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}\

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$\\log_a(\\var{(x1[4])*y1})-\\log_a(\\var{x1[4]})=\\log_a($ [[0]]$)$

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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}\

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Use laws for addition and subtraction of logarithms to simplify a given logarithmic expression to an arbitrary base.

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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.

#### 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}\

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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]]$)$

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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]] $\\log_a(\\var{x1[5]})$

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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]]$\\log_a(\\var{x1[5]})$

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

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Changing the subject of an equation involving logarithms often requires the use of the equivalence

\n

\$\\log_ba=c \\Longleftrightarrow a=b^c\\text{.}\$

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Rearrange the equation to find $x$.

\n

$\\log_\\var{f}(x)=\\var{f1}$

\n

$x=$ [[0]]

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Make $x$ the subject of the following equation.

\n

$\\log_\\var{g1}(x)=y+\\var{g2}$

\n

$x=$ [[0]]

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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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$\\log_a(a^x)$

", "

$a^{\\log_a(x)}$

", "

$e^{\\ln(x)}$

", "

$\\log_{10}(x)$

", "

$\\log_e(x)$

", "

$\\ln(e^x)$

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Which of the following expressions are equivalent to $x$?

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Rearrange some expressions involving logarithms by applying the relation $\\log_b(a) = c \\iff a = b^c$.

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#### a)

\n

i)

\n

We can rearrange logarithms using indices.

\n

\$\\log_ba=c \\Longleftrightarrow a=b^c\$

\n

Using 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

\n

#### b)

\n

i)

\n

We can use the equivalence to rewrite our equation.

\n

\$\\log_ba=c \\Longleftrightarrow a=b^c\$

\n

We can write out our values to makes it easier.

\n

\\\begin{align} a&=x \\\\ b&=\\var{g1}\\\\ c&=y+\\var{g2} \\end{align}\

\n

Then we can write out our equation in the required form.

\n

\$x=\\var{g1}^{y+\\var{g2}}\$

\n

\n

#### c)

\n

We can use the same equivalence as in part b)

\n

\$\\log_ba=c \\Longleftrightarrow a=b^c\$

\n

We 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}

\n

\n

#### d)

\n

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}\

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Apply and combine logarithm laws in a given equation to find the value of $x$.

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#### a)

\n

We can use the logarithm law

\n

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

\n

to 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}\

\n

This means we can write our expression as

\n

\$\\log_\\var{b1}(x-\\var{b2})+\\log_\\var{b1}({x})=\\var{b4}\\text{.}\$

\n

Then using the rule

\n

\$\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\text{,}\$

\n

we 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}\

\n

We then rely on the definition of $\\log_a$

\n

\$b=a^c \\Longleftrightarrow \\log_{a}b=c\$

\n

to write our equation as

\n

\\\begin{align} x^2-\\var{b2}x&=\\var{b1}^\\var{b4}\\\\ &=\\var{b1^b4}\\text{.} \\end{align}\

\n

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}\

\n

As logarithms can only be applied to positive numbers, the only possible value for $x$ is $\\var{b}$.

\n

#### b)

\n

$\\ln(x)$ is a shorthand for $\\log_e(x)$, so we can apply the same laws of logarithms here.

\n

Therefore applying the rule

\n

\$k\\log_a(x)=\\log_a(x^k)\$

\n

we can write our equation as

\n

\$\\ln(x^\\var{p})+\\ln(\\var{q})=\\var{m}\\text{.}\$

\n

Then using the rule

\n

\$\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\$

\n

we can write our equation as

\n

\$\\ln(\\var{q}x^\\var{p})=\\var{m}\\text{.}\$

\n

As $\\ln=\\log_e$ we can use

\n

\$a=b^c \\Longleftrightarrow \\log_ba=c\$

\n

to write our equation as

\n

\$\\var{q}x^\\var{p}=e^\\var{m}\\text{.}\$

\n

We 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}\

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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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You may find the following conversion useful

\n

\$\\ln(x)=\\log_e(x)\$

", "variableReplacements": [], "showFeedbackIcon": true, "marks": 0}], "prompt": "

Solve for $x$ and leave your answer in the form  $x=\\displaystyle\\frac{e^{a}}{b}$.

\n

$\\var{p}\\ln(x)+\\ln(\\var{q})=\\var{m}$

\n

$x=$ [[0]]

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