Question

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

• A. $10\cdot 5$
• B. $11\cdot 5$
• C. $12\cdot 5$
• D. $13\cdot 5$

Answer 1

Answer: (C)

$49^{1-{\log }_{7}2}+5^{-{\log }_{5}4}$
$=49^{{\log }_{7}7-{\log }_{7}2}+\frac{1}{5^{{\log }_{5}4}}$
$=49^{{\log }_7\left(\frac{7}{2}\right)}+\frac{1}{4}$
$=\frac{49}{4}+\frac{1}{4}=\frac{50}{4}=\frac{25}{2}=12.5$