Apply and combine logarithm laws in a given equation to find the value of $x$.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Solve for $x$
", "advice": "We can use the logarithm law
\n\\[k\\log_a(x)=\\log_a(x^k)\\text{,}\\]
\nto 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}\\]
This means we can write our expression as
\n\\[\\log_\\var{b1}(x-\\var{b2})+\\log_\\var{b1}({x})=\\var{b4}\\text{.}\\]
\nThen using the rule
\n\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\text{,}\\]
\nwe 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}\\]
We then rely on the definition of $\\log_a$
\n\\[b=a^c \\Longleftrightarrow \\log_{a}b=c\\]
\nto write our equation as
\n\\[\\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;
\n\\[\\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 $\\var{b}$.
\n$\\ln(x)$ is a shorthand for $\\log_e(x)$, so we can apply the same laws of logarithms here.
\nTherefore applying the rule
\n\\[k\\log_a(x)=\\log_a(x^k)\\]
\nwe can write our equation as
\n\\[\\ln(x^\\var{p})+\\ln(\\var{q})=\\var{m}\\text{.}\\]
\nThen using the rule
\n\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\]
\nwe can write our equation as
\n\\[\\ln(\\var{q}x^\\var{p})=\\var{m}\\text{.}\\]
\nAs $\\ln=\\log_e$ we can use
\n\\[a=b^c \\Longleftrightarrow \\log_ba=c\\]
\nto write our equation as
\n\\[\\var{q}x^\\var{p}=e^\\var{m}\\text{.}\\]
\nWe 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}\\]
$\\log_\\var{b1}(x-\\var{b2})-\\log_\\var{b1}\\left(\\displaystyle\\frac{1}{x}\\right)=\\var{b4}$
\n$x=$ [[0]]
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\n$\\var{p}\\ln(x)+\\ln(\\var{q})=\\var{m}$
\n$x=$ [[0]]
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\n\\[\\ln(x)=\\log_e(x)\\]
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