# Please, write the logarithm as a ratio of common logarithms

Please, write the logarithm as a ratio of common logarithms and natural logarithms.
$$\displaystyle{{\log}_{{9}}{\left({69}\right)}}$$
a) common logarithms
b) natural logarithms

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Given, $$\displaystyle{{\log}_{{9}}{\left({69}\right)}}$$
a) let $$\displaystyle{{\log}_{{b}}{a}}={x}$$ then we know that $$\displaystyle{b}^{{x}}={a}$$
Now taking common log (to the base 10) on both side, we get
$$\displaystyle{{\log{{b}}}^{{x}}=}{\log{{a}}}$$
$$\displaystyle{x}{\log{{b}}}={\log{{a}}}$$
$$\displaystyle{x}={\frac{{{\log{{a}}}}}{{{\log{{b}}}}}}$$
As $$\displaystyle{x}={{\log}_{{b}}{a}}$$, we can state $$\displaystyle{{\log}_{{b}}{a}}={\frac{{{\log{{a}}}}}{{{\log{{b}}}}}}$$
Hence, in common
$$\displaystyle{{\log}_{{9}}{\left({69}\right)}}={\frac{{{\log{{69}}}}}{{{\log{{9}}}}}}$$
b) $$\displaystyle{b}^{{x}}={a}$$
Taking natural logarithsm we get
$$\displaystyle{{\ln{{b}}}^{{x}}=}{\ln{{a}}}---{x}{\ln{{b}}}={\ln{{a}}}$$
$$\displaystyle{x}={\frac{{{\ln{{a}}}}}{{{\ln{{b}}}}}}$$
In natural logarithm $$\displaystyle{{\log}_{{9}}{\left({69}\right)}}={\frac{{{\ln{{69}}}}}{{{\ln{{9}}}}}}$$