Question

If $\log 4=0.6020$, then find the value of $\log 80$.

Answer 1

Calculate the required logarithmic value

It is given that $\log 4=0.6020$.

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

So, the value of $\log 80$ can be calculated as,

$\log 80={\log }_{10}80$

$\Rightarrow \log 80={\log }_{10}\left(2\times 4\times 10\right)$ $\left[\because 80=2\times 4\times 10\right]$

$\Rightarrow \log 80={\log }_{10}2+{\log }_{10}4+{\log }_{10}10$ $\mathrm{[}\mathrm{\because }\mathrm{}\log \left(m\times n\right)\mathrm{=}\mathrm{logm}\mathrm{+}\mathrm{logn}\mathrm{]}$

$\Rightarrow \log 80=\frac{1}{2}\times 2{\log }_{10}2+{\log }_{10}4+{\log }_{10}10$

$\Rightarrow \log 80=\frac{1}{2}\times {\log }_{10}{2}^2+{\log }_{10}4+{\log }_{10}10$ $\mathrm{[}\mathrm{\because }\mathrm{}\mathrm{nlogm}\mathrm{=}{\mathrm{logm}}^{\mathrm{n}}\mathrm{]}$

$\Rightarrow \log 80=\frac{1}{2}\times {\log }_{10}4+{\log }_{10}4+{\log }_{10}10$

$\Rightarrow \log 80=\frac{1}{2}\times 0.6020+0.6020+{\log }_{10}10$ $\left[\because \log 4=0.6020\right]$

$\Rightarrow \log 80=0.3010+0.6020+1$ $\mathrm{[}\mathrm{\because }\mathrm{}{\log }_{\mathrm{10}}\mathrm{10}\mathrm{=}\mathrm{1}\mathrm{]}$

$\Rightarrow \log 80=1.9030$

Hence, the required value of $\log 80$ is $1.9030$.