Question

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

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

Answer 1

Answer: (C)

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

Answer 2

Given, $\log _{7} 2=m$
$\therefore \log _{49}(28) =\log _{7^{2}}\left(2^{2} \times 7\right)$
$=\frac{1}{2}\left[2 \log _{7} 4+\log _{7} 7\right]$
$=\frac{1}{2} \log _{7} 2^2+\frac{1}{2}1$

$\frac{2}{2} log_72+\frac12$

m+½

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