Question

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

• A. $\frac{25}{2}$
• B. $\frac{49}{4}$
• C. $12$
• D. none of these

Answer 1

The correct option is A $\frac{25}{2}$

${49}^A={49}^{1-{\log }_{7}2}=\frac{49}{{49}^{{\log }_{7}2}}=\frac{49}{7^{{\log }_{7}4}}=\frac{49}{4}[\because a^{{\log }_{a}b}=b]$
$5^B=5^{-{\log }_{5}4}=5^{{\log }_5\frac{1}{4}}=\frac{1}{4}$
$\therefore {49}^A+5^B=\frac{25}{2}$.