Question

The value of $(\log m+\log m^2+\log m^3+...+\log m^n)$ is equal to

• A. $\frac{n(n+1)}{2}$
• B. $\frac{mn}{2}$
• C. $\frac{n(n+1)}{2}\log m$
• D. $n(n+1)\cdot \log m^2$

Answer 1

Answer: (C)

$\frac{n(n+1)}{2}\log m$

$\log m+\log m^2+\log m^3+...+\log m^n$
$=\log (m.m^2.m^3...m^n)$
$=\log m^{(1+2+3+...n)}=\log m^{\frac{n(n+1)}{2}}$

$=\frac{n(n+1)}{2}\log m$