Sarvesh K Verma Solutions for Chapter: Logarithm, Exercise 1: Introductory Exercise

The value of ${\log }_{10}2+16{\log }_{10}\left(\frac{16}{15}\right)+12{\log }_{10}\left(\frac{25}{24}\right)+7{\log }_{10}\left(\frac{81}{80}\right)$ is :

If ${\log }_{10}\mathrm{x}=\mathrm{a},{\log }_{10}\mathrm{y}=\mathrm{b}$ and ${\log }_{10}\mathrm{z}=\mathrm{c}$, then antilog $(\mathrm{p} \mathrm{a}+\mathrm{q} \mathrm{b}-\mathrm{r} \mathrm{c})=?$

${\log }_{10}{\mathrm{a}}^{\mathrm{p}}·{\mathrm{b}}^{\mathrm{q}}·{\mathrm{c}}^{\mathrm{r}}=?$

If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in GP then ${\log }_{10}\mathrm{a},{\log }_{10}\mathrm{b},{\log }_{10}\mathrm{c}$ are in :

If ${\log }_{10}\mathrm{x},{\log }_{10}\mathrm{y},{\log }_{10}\mathrm{z}$ are in AP then $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in :

If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in GP then $\frac{1}{{\log }_{\mathrm{a}}\mathrm{x}},\frac{1}{{\log }_{\mathrm{b}}\mathrm{x}},\frac{1}{{\log }_{\mathrm{c}}\mathrm{x}}$ are in :

If $\log (\mathrm{x}-1)+\log (\mathrm{x}+1)=3\log 2$, then $\mathrm{x}$ is equal to :

If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are three consecutive integers, then $\log (\mathrm{ac}+1)$ has the value :