Alvin Parks

2022-11-23

Solve x in logarithm equation
I am trying to solve $x$ for $2lo{g}_{10}\left(x-4\right)-lo{g}_{10}4\left(x-1\right)=0$
I have the key with the answer 10 and have confirmed this is correct using Wolfram Alpha but which steps should I take to reach that answer?

Expert

Use the properties of logarithms
$k{\mathrm{log}}_{a}x={\mathrm{log}}_{a}\left({x}^{k}\right)$
${\mathrm{log}}_{a}x-{\mathrm{log}}_{a}y={\mathrm{log}}_{a}\frac{x}{y}$
${\mathrm{log}}_{a}x=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x=1$

Uriah Molina

Expert

$2{\mathrm{log}}_{10}\left(x-4\right)-{\mathrm{log}}_{10}4\left(x-1\right)=0$
A rule of logarithm: $2{\mathrm{log}}_{10}\left(x-4\right)={\mathrm{log}}_{10}\left(x-4{\right)}^{2}$
${\mathrm{log}}_{10}\left(x-4{\right)}^{2}-{\mathrm{log}}_{10}\left(4x-4\right)=0$
Another rule of logarithm: Express as the left side of the equation as a single product:
${\mathrm{log}}_{10}\frac{\left(x-4{\right)}^{2}}{4x-4}=0$
Convert to exponential form:
${10}^{{\mathrm{log}}_{10}\frac{\left(x-4{\right)}^{2}}{4x-4}}={10}^{0}$
or
$\frac{\left(x-4{\right)}^{2}}{4x-4}=1$
and when algebraically rearranging
${x}^{2}-8x+16=4x-4$
or
${x}^{2}-12x+20=0$
which is now a quadratic equation that you can solve.

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