Terrence Moore

2022-05-01

Simplify a logarithm function

$2\mathrm{log}\sqrt[4]{10}-\mathrm{ln}{e}^{-7}+{\mathrm{log}}_{9}\sqrt{3}$

I want to simplify this function. I believe that $2\mathrm{log}\sqrt[4]{10}$ can become $\mathrm{log}\sqrt{10}$ but now I'm stuck.

Is it possible that $\mathrm{ln}e}^{-7$ can be just $-7$?

eldgamliru9x

Beginner2022-05-02Added 16 answers

$\mathrm{log}\sqrt[4]{10}=\frac{1}{4}\mathrm{log}10$

${\mathrm{ln}e}^{-7}=-7\mathrm{ln}e=-7\cdot 1$

$\mathrm{log}}_{9}\sqrt{3}=\frac{1}{2}{\mathrm{log}}_{9}3=\frac{1}{2}{\mathrm{log}}_{9}\sqrt{9}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4$

Probably "$\mathrm{log}10$" is intended to mean ${\mathrm{log}}_{10}10$, so that is equal to 1.

I disapprove of using "$\mathrm{log}$" with no base to mean $\mathrm{log}}_{10$ when there's no special context saying that's the right base to use.

Cristal Roth

Beginner2022-05-03Added 13 answers

$2\mathrm{log}\sqrt[4]{10}-\mathrm{ln}{e}^{-7}+{\mathrm{log}}_{9}\sqrt{3}=\frac{2}{4}\mathrm{log}10-(-7)+\frac{1}{2}{\mathrm{log}}_{9}3=\frac{1}{2}+7+\frac{1}{4}\dots $

Properties used:

${\mathrm{log}}_{a}{a}^{n}=n,\text{}\text{}\text{}\text{}\text{}\text{}{\mathrm{log}}_{a}{x}^{n}=n{\mathrm{log}}_{a}x$

and, of course, the very definition of logarithm and the assumed fact that you surely meant $\mathrm{log}={\mathrm{log}}_{10}$

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