Question

If ${\log }_{7}2=\lambda$, then the value of ${\log }_{49}\left(28\right)$ is

• A. $(2\lambda +1)$
• B. $(2\lambda +3)$
• C. $\frac{1}{2}(2\lambda +1)$
• D. $2(2\lambda +1)$

Answer 1

Answer: (C)

Given, ${\log }_{7}2=\lambda$
$\therefore {\log }_{49}(28)={\log }_{7^2}\left(2^2\times 7\right)$
$=\frac{1}{2}\left[2{\log }_{7}2+{\log }_{7}7\right]$
$=\frac{1}{2}(2\lambda +1)$