Matilda Fox

2022-07-18

Expansion of Logarithms with Cube Roots
Does the following expand to the following
${\mathrm{log}}_{6}\left({11}^{6}\sqrt[3]{12}\right)$
= $6{\mathrm{log}}_{6}\left(11\right)+{\mathrm{log}}_{6}\left(\sqrt[3]{12}\right)$

kartonaun

Yes and this is equal to
$6{\mathrm{log}}_{6}\left(11\right)+\frac{1}{3}\left(\mathrm{log}\left(2\right)+1\right)$

Bruno Thompson

First note that
$\sqrt[n]{x}={x}^{\frac{1}{n}}$
${\mathrm{log}}_{b}\left(xy\right)={\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$
${\mathrm{log}}_{b}\left({x}^{n}\right)=n{\mathrm{log}}_{b}\left(x\right)$
${\mathrm{log}}_{b}\left(x\right)=\frac{{\mathrm{log}}_{10}\left(x\right)}{{\mathrm{log}}_{10}\left(b\right)}=\frac{\mathrm{log}\left(x\right)}{\mathrm{log}\left(b\right)}$
So then
${\mathrm{log}}_{6}\left({11}^{6}\cdot \sqrt[3]{12}\right)={\mathrm{log}}_{6}\left({11}^{6}\cdot {12}^{\frac{1}{3}}\right)={\mathrm{log}}_{6}\left({11}^{6}\right)+{\mathrm{log}}_{6}\left({12}^{\frac{1}{3}}\right)=6{\mathrm{log}}_{6}\left(11\right)+\frac{1}{3}{\mathrm{log}}_{6}\left(12\right)$
Or perhaps
$\frac{6\mathrm{log}\left(11\right)}{\mathrm{log}\left(6\right)}+\frac{\mathrm{log}\left(12\right)}{3\mathrm{log}\left(6\right)}$

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