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Solve for $x$.

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

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$x=$ [[0]]

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

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

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$x=$ [[0]]

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You may find the following conversion useful

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

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

"}, "functions": {}, "advice": "

#### a)

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We can use the logarithm law

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

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to also give a more specific rule

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\\\begin{align} \\log_a\\left(\\frac{1}{x}\\right)&=\\log_a(x^{-1})\\\\ &=-\\log_a(x)\\text{.} \\end{align}\

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This means we can write our expression as

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

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Then using the rule

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

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we can write our equation as

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

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

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

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to write our equation as

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

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We can then write out our equation and solve either by factorising or using the quadratic formula;

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\\\begin{align} x^2-\\var{b2}x-\\var{b1^{b4}}&=0\\\\ (x+2)(x-\\var{b})&=0\\text{.} \\end{align}\

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

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#### b)

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

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Therefore applying the rule

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

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we can write our equation as

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

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Then using the rule

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

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we can write our equation as

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

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

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

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to write our equation as

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

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We then just need to rearrange our equation

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\\\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}\

", "tags": ["logarithm", "Logarithm", "logs", "Logs", "taxonomy"], "type": "question", "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/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "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/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}