Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}(2\lambda +1)$

Given ${\log }_{7}2=\lambda$

$\therefore {\log }_{49}\left(28\right)={\log }_{7^2}(2^2\times 7)$

$=\frac{1}{2}[2{\log }_{7}2+{\log }_{7}7]$

$=\frac{1}{2}(2\lambda +1)$