1. Tardigrade
  2. Question
  3. Mathematics
  4. If log 2 p+ log 8 p+ log 32 p=(46/5), then p=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $\log _2 p+\log _8 p+\log _{32} p=\frac{46}{5}$, then $p=$

Logarithm

Solution:

Given, $\log _2 p+\log _8 p+\log _{32} p=\frac{46}{5}$
$\log _2 p+(\frac{1}{3}) \log _2 p+(\frac{1}{5}) \log _2 p=\frac{46}{5}$
$\log _2 p(1+\frac{1}{3}+\frac{1}{5})=\frac{46}{5}$
$(\frac{15+5+3}{15}) \log _2 p=\frac{46}{5}$
$(\frac{23}{15}) \log _2 p=\frac{46}{5}$
$\log _2 p=\frac{46}{5} \times \frac{15}{23}$
$\log _2 p=6$ $p=2^6=64$.