burubukuamaw

2022-03-25

Condense using properties of logarithms: $\mathrm{log}13+\frac{13}{\mathrm{log}\left(x-9\right)}–3\mathrm{log}\left(x+9\right)$

Aidyn Wall

Given :
$\mathrm{log}13+\frac{13}{\mathrm{log}\left(x-9\right)}-3\mathrm{log}\left(x+9\right)$
Using ${\mathrm{log}a}^{b}=b\mathrm{log}a,$
$\mathrm{log}13+\frac{13}{\mathrm{log}\left(x-9\right)}-3\mathrm{log}\left(x+9\right)=\mathrm{log}13+{\mathrm{log}\left(x-9\right)}^{\frac{13}{-}}{\mathrm{log}\left(x+9\right)}^{3}$
Using $\mathrm{log}a+\mathrm{log}b=\mathrm{log}\left(ab\right)$,
$\mathrm{log}13+\frac{1}{3}\mathrm{log}\left(x-9\right)-3\mathrm{log}\left(x+9\right)=\mathrm{log}\left(13\left(x-9{\right)}^{\frac{1}{3}}\right)-\mathrm{log}\left(x+9{\right)}^{3}$
Using $\mathrm{log}a-\mathrm{log}b=\mathrm{log}\left(\frac{a}{b}\right)$,
$\mathrm{log}13+\frac{1}{3}\mathrm{log}\left(x-9\right)-3\mathrm{log}\left(x+9\right)=\mathrm{log}\left(\frac{13\left(x-9{\right)}^{\frac{1}{3}}}{\left(x+9{\right)}^{3}}\right)$
Therefore
$\mathrm{log}13+\frac{1}{3}\mathrm{log}\left(x-9\right)-3\mathrm{log}\left(x+9\right)=\mathrm{log}\left(\frac{13\left(x-9{\right)}^{\frac{1}{3}}}{\left(x+9{\right)}^{3}}\right)$

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