Question

If ${\log }_{2}p+{\log }_{8}p+{\log }_{32}p=\frac{46}{5}$, then $p=$

• A. 128
• B. 64
• C. 32
• D. 256

Answer 1

Answer: (B)

Given, ${\log }_{2}p+{\log }_{8}p+{\log }_{32}p=\frac{46}{5}$
${\log }_{2}p+(\frac{1}{3}){\log }_{2}p+(\frac{1}{5}){\log }_{2}p=\frac{46}{5}$
${\log }_{2}p(1+\frac{1}{3}+\frac{1}{5})=\frac{46}{5}$
$(\frac{15+5+3}{15}){\log }_{2}p=\frac{46}{5}$
$(\frac{23}{15}){\log }_{2}p=\frac{46}{5}$
${\log }_{2}p=\frac{46}{5}\times \frac{15}{23}$
${\log }_{2}p=6$ $p=2^6=64$.