# Solve the equation. log_4(8x) - log_4(x - 1)= 2

Solve the equation.
${\mathrm{log}}_{4}\left(8x\right)-{\mathrm{log}}_{4}\left(x-1\right)=2$
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Izabelle Frost
${\mathrm{log}}_{4}\left(8x\right)-{\mathrm{log}}_{4}\left(x-1\right)=2$ remember $\mathrm{log}\left(a/b\right)=\mathrm{log}\left(a\right)-\mathrm{log}\left(b\right)$
${\mathrm{log}}_{4}\left(\frac{8x}{x-1}\right)={\mathrm{log}}_{4}{4}^{2}{\mathrm{log}}_{b}\left({b}^{m}\right)=m{\mathrm{log}}_{b}\left(b\right)=m$
$\frac{8x}{x-1}=16$
8x= 16x -16
8x=16
x=2
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yasusar0
Solve the equation.
${\mathrm{log}}_{4}\left(8x\right)-{\mathrm{log}}_{4}\left(x-1\right)=2$
Use the law of logarithm: ${\mathrm{log}}_{a}\left(\frac{A}{B}\right)={\mathrm{log}}_{a}A-{\mathrm{log}}_{a}B$, to combine the logarithms.
${\mathrm{log}}_{4}\left(\frac{8x}{x-1}\right)=2$
Use the definition of the logarithmic function: ${\mathrm{log}}_{a}x=y⇔{a}^{y}=x$, to put the equation in exponential form.
${4}^{2}=\frac{8x}{x-1}$
$16=\frac{8x}{x-1}$
Now solve for x.
$16\left(x-1\right)=8x←$ multiply both sides by (x-1).
16x-16=8x
16x-8x=16
$8x=16\to x=\frac{16}{8}$
So, x=2.
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