Question

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

Answer 1

Given,

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

${\log }_{2}p+\left(\frac{1}{3}\right){\log }_{2}p+\left(\frac{1}{5}\right){\log }_{2}p=\frac{46}{5}$

${\log }_{2}p(1+\frac{1}{3}+\frac{1}{5})=\frac{46}{5}$

$\left(\frac{15+5+3}{15}\right){\log }_{2}p=\frac{46}{5}$

$\left(\frac{23}{15}\right){\log }_{2}p=\frac{46}{5}$

${\log }_{2}p=\frac{46}{5}\times \frac{15}{23}$

${\log }_{2}p=6$

$p=2^6=64$.