waigaK

2021-08-14

Please, write the logarithm as a ratio of common logarithms and natural logarithms.

${\mathrm{log}}_{9}\left(69\right)$

a) common logarithms

b) natural logarithms

a) common logarithms

b) natural logarithms

SkladanH

Skilled2021-08-15Added 80 answers

Given, ${\mathrm{log}}_{9}\left(69\right)$

a) let${\mathrm{log}}_{b}a=x$ then we know that ${b}^{x}=a$

Now taking common log (to the base 10) on both side, we get

${\mathrm{log}b}^{x}=\mathrm{log}a$

$x\mathrm{log}b=\mathrm{log}a$

$x=\frac{\mathrm{log}a}{\mathrm{log}b}$

As$x={\mathrm{log}}_{b}a$ , we can state $\mathrm{log}}_{b}a=\frac{\mathrm{log}a}{\mathrm{log}b$

Hence, in common

$\mathrm{log}}_{9}\left(69\right)=\frac{\mathrm{log}69}{\mathrm{log}9$

b)${b}^{x}=a$

Taking natural logarithsm we get

${\mathrm{ln}b}^{x}=\mathrm{ln}a---x\mathrm{ln}b=\mathrm{ln}a$

$x=\frac{\mathrm{ln}a}{\mathrm{ln}b}$

In natural logarithm$\mathrm{log}}_{9}\left(69\right)=\frac{\mathrm{ln}69}{\mathrm{ln}9$

a) let

Now taking common log (to the base 10) on both side, we get

As

Hence, in common

b)

Taking natural logarithsm we get

In natural logarithm

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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