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

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a)

We can use the logarithm law

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

to also give a more specific rule

\\[\\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

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

Then using the rule

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

we can write our equation as

\\[\\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 $\\log_a$

\\[b=a^c \\Longleftrightarrow \\log_{a}b=c\\]

to write our equation as

\\[\\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;

\\[\\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 $\\var{b}$.

b)

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

Therefore applying the rule

\\[k\\log_a(x)=\\log_a(x^k)\\]

we can write our equation as

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

Then using the rule

\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\]

we can write our equation as

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

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

\\[a=b^c \\Longleftrightarrow \\log_ba=c\\]

to write our equation as

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

We then just need to rearrange our equation

\\[\\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 $x$.

$\\log_\\var{b1}(x-\\var{b2})-\\log_\\var{b1}\\left(\\displaystyle\\frac{1}{x}\\right)=\\var{b4}$

$x=$ [[0]]

You may find the following conversion useful

\\[\\ln(x)=\\log_e(x)\\]

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

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

$x=$ [[0]]

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