Question

The value of $2({49}^A+5^B)$, where $A=1-{\log }_{7}2$ and $B=-{\log }_{5}4$ is

Answer 1

$A=1-{\log }_{7}2={\log }_{7}7-{\log }_{7}2[\because {\log }_{a}a=1]={\log }_7\frac{7}{2}[\because \log a-\log b=\log \frac{a}{b}]$
$\Rightarrow \frac{7}{2}=7^A$
$\Rightarrow \frac{49}{4}={49}^A$ ...(1)
$B=-{\log }_{5}4={\log }_5\frac{1}{4}[\because x\log a=\log a^x]$
$\Rightarrow \frac{1}{4}=B^5$ ...(2)
Adding (1) and (2), we get
$2({49}^A+B^5)=\frac{49+1}{2}=25$

Ans: 25