Question

If ${\log }_{7}2=a$, then ${\log }_{49}28$ is equal to

• A. $\frac{1+2a}{4}$
• B. $\frac{1+2a}{2}$
• C. $\frac{1+2a}{3}$
• D. none of these

Answer 1

Answer: (B)

$\frac{1+2a}{2}$

We have,
${\log }_{49}28=x$
Taking exponential
${49}^x=28$
$7^{2x}=2^2{7}^1$
Taking $\log$ on both sides, we get
$2x\log 7=2\log 2+\log 7$
$\therefore x=\frac{2\log 2+\log 7}{2\log 7}$
$\therefore x=\frac{2\log 2}{2\log 7}+\frac{\log 7}{2\log 7}$
$\therefore x={\log }_{7}2+\frac{1}{2},[\because {\log }_{a}b=\frac{\log b}{\log a}]$
$\therefore x=a+\frac{1}{2}=\frac{2a+1}{2}$ $(\because {\log }_{7}2=a)$