Question

If $\log 2=0.3010$ and $\log 3=0.4771$ find the value of $\log 25$?

Answer 1

Calculate the required logarithmic value

It is given that $\log 2=0.3010$

And, $\log 3=0.4771$

As we know the common logarithm is treated as the logarithm with base $10$.

So, the value of $\log 25$ can be calculated as,

$\log 25={\log }_{10}25$

$\Rightarrow \log 25={\log }_{10}\left(5\times 5\right)$ $\left[\because 25=5\times 5\right]$

$\Rightarrow \log 25={\log }_{10}5+{\log }_{10}5$ $\mathrm{[}\mathrm{\because }\mathrm{}\log \mathrm{(}\mathrm{m}\mathrm{\times }\mathrm{n}\mathrm{)}\mathrm{=}\mathrm{logm}\mathrm{+}\mathrm{logn}\mathrm{]}$

$\Rightarrow \log 25={\log }_{10}\left(\frac{10}{2}\right)+{\log }_{10}\left(\frac{10}{2}\right)$ $\left[\because 5=\frac{10}{2}\right]$

$\Rightarrow \log 25={\log }_{10}10-{\log }_{10}2+{\log }_{10}10-{\log }_{10}2$ $\mathrm{[}\mathrm{\because }\mathrm{}\log \mathrm{\left(}\frac{\mathrm{m}}{\mathrm{n}}\mathrm{\right)}\mathrm{=}\mathrm{logm}\mathrm{-}\mathrm{logn}\mathrm{]}$

$\Rightarrow \log 25=1-{\log }_{10}2+1-{\log }_{10}2$ $\mathrm{[}\mathrm{\because }\mathrm{}{\log }_{\mathrm{10}}\mathrm{10}\mathrm{=}\mathrm{1}\mathrm{]}$

$\Rightarrow \log 25=1-0.3010+1-0.3010$ $\left[\because \log 2=0.3010\right]$

$\Rightarrow \log 25=2-0.6020$

$\Rightarrow \log 25=1.3980$

Hence, the required value of $\log 25$ is $1.3980$.