Matilda Fox

2022-07-18

Expansion of Logarithms with Cube Roots

Does the following expand to the following

${\mathrm{log}}_{6}({11}^{6}\sqrt[3]{12})$

= $6{\mathrm{log}}_{6}(11)+{\mathrm{log}}_{6}(\sqrt[3]{12})$

Does the following expand to the following

${\mathrm{log}}_{6}({11}^{6}\sqrt[3]{12})$

= $6{\mathrm{log}}_{6}(11)+{\mathrm{log}}_{6}(\sqrt[3]{12})$

kartonaun

Beginner2022-07-19Added 14 answers

Yes and this is equal to

$6{\mathrm{log}}_{6}(11)+\frac{1}{3}(\mathrm{log}(2)+1)$

$6{\mathrm{log}}_{6}(11)+\frac{1}{3}(\mathrm{log}(2)+1)$

Bruno Thompson

Beginner2022-07-20Added 9 answers

First note that

$\sqrt[n]{x}={x}^{\frac{1}{n}}$

${\mathrm{log}}_{b}(xy)={\mathrm{log}}_{b}(x)+{\mathrm{log}}_{b}(y)$

${\mathrm{log}}_{b}({x}^{n})=n{\mathrm{log}}_{b}(x)$

${\mathrm{log}}_{b}(x)=\frac{{\mathrm{log}}_{10}(x)}{{\mathrm{log}}_{10}(b)}=\frac{\mathrm{log}(x)}{\mathrm{log}(b)}$

So then

${\mathrm{log}}_{6}({11}^{6}\cdot \sqrt[3]{12})={\mathrm{log}}_{6}({11}^{6}\cdot {12}^{\frac{1}{3}})={\mathrm{log}}_{6}({11}^{6})+{\mathrm{log}}_{6}({12}^{\frac{1}{3}})=6{\mathrm{log}}_{6}(11)+\frac{1}{3}{\mathrm{log}}_{6}(12)$

Or perhaps

$\frac{6\mathrm{log}(11)}{\mathrm{log}(6)}+\frac{\mathrm{log}(12)}{3\mathrm{log}(6)}$

$\sqrt[n]{x}={x}^{\frac{1}{n}}$

${\mathrm{log}}_{b}(xy)={\mathrm{log}}_{b}(x)+{\mathrm{log}}_{b}(y)$

${\mathrm{log}}_{b}({x}^{n})=n{\mathrm{log}}_{b}(x)$

${\mathrm{log}}_{b}(x)=\frac{{\mathrm{log}}_{10}(x)}{{\mathrm{log}}_{10}(b)}=\frac{\mathrm{log}(x)}{\mathrm{log}(b)}$

So then

${\mathrm{log}}_{6}({11}^{6}\cdot \sqrt[3]{12})={\mathrm{log}}_{6}({11}^{6}\cdot {12}^{\frac{1}{3}})={\mathrm{log}}_{6}({11}^{6})+{\mathrm{log}}_{6}({12}^{\frac{1}{3}})=6{\mathrm{log}}_{6}(11)+\frac{1}{3}{\mathrm{log}}_{6}(12)$

Or perhaps

$\frac{6\mathrm{log}(11)}{\mathrm{log}(6)}+\frac{\mathrm{log}(12)}{3\mathrm{log}(6)}$

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