# log3(1/9)

Question
Logarithms
$$\displaystyle{\log{{3}}}{\left(\frac{{1}}{{9}}\right)}$$

2020-11-28
$$\displaystyle{\log{{3}}}{\left(\frac{{1}}{{9}}\right)}={\log{{3}}}{\left({3}^{{-{{2}}}}\right)}$$ Rewrite $$\displaystyle\frac{{1}}{{9}}$$ as a power of 3. =-2 Use the rule $$\displaystyle{\log{{\mathbf{^}}}}{x}={x}.$$ -2.

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