Question

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

• A. $2\left(1+2m\right)$
• B. $\frac{1+2m}{2}$
• C. $\frac{2}{1+2m}$
• D. $1+m$

Answer 1

The correct option is B $\frac{1+2m}{2}$

We have,

${\log }_{7}2=m$ $........\left(1\right)$

Since,

${\log }_{49}28$

$\Rightarrow {\log }_{7^2}28$

$\Rightarrow \frac{1}{2}{\log }_{7}28$ $\because {\log }_{a^n}b=\frac{1}{n}{\log }_a=b$

$\Rightarrow \frac{1}{2}{\log }_7(4\times 7)$

$\Rightarrow \frac{1}{2}{\log }_{7}4+\frac{1}{2}{\log }_{7}7$ $\because {\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$

$\Rightarrow \frac{1}{2}{\log }_7{2}^2+\frac{1}{2}\times 1$

$\Rightarrow \frac{2}{2}{\log }_{7}2+\frac{1}{2}$ $\because {\log }_a{m}^n=n{\log }_{a}m$

$\Rightarrow {\log }_{7}2+\frac{1}{2}$

$\Rightarrow m+\frac{1}{2}$ From $equation\left(1\right)$

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

Hence, this is the answer.