Please, write the logarithm as a ratio of common logarithms

waigaK 2021-08-14 Answered
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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Expert Answer

SkladanH
Answered 2021-08-15 Author has 17716 answers
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 \(\displaystyle{{\log}_{{9}}{\left({69}\right)}}={\frac{{{\ln{{69}}}}}{{{\ln{{9}}}}}}\)
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