Question

If $a=1-{\log }_{7}2$ and $b={\log }_5\left(\frac{1}{4}\right)$ then the value of $(49{)}^a+5^b$ equals

• A. $\frac{21}{2}$
• B. $\frac{23}{2}$
• C. $\frac{25}{2}$
• D. $\frac{29}{2}$

Answer 1

Answer: (C)

Given, $a=1-{\log }_{7}2,b={\log }_5\left(\frac{1}{4}\right)$
Now, Let $E=49^a+5^b=49^{1-{\log }_{7}2}+5^{{\log }_5\left(\frac{1}{4}\right)}=49\cdot 7^{-{\log }_{7}4}+\frac{1}{4}=\frac{49}{4}+\frac{1}{4}=\frac{50}{4}=\frac{25}{2}$