Question

Logx base 2 + logx base 3 = 5

Answer 1

use change of base

\[\log _{2} x = \log _{e}x/\log _{e}2\]

\[\log _{3} x = \log _{e} x/\log _{e}3\]

so the problem is now

\[\log _{e} x/\log _{e} 2 + \log _{e}x/\log _{e}3 = 5\]

\[\log _{e} x(1/ \log _{e}2 +1/\log _{e}3) = 5\]

\[\log _{e} x = 5/(1/\log _{e}2 + 1/\log _{e}3)\]

get a common denominator for the fraction

\[\log _{x} = 5/(\log _{e} 3 + \log _{e}2)/(\log _{e}2timeslog _{e}3))\]

dividing by a fraction, invert and multiply

\[\log _{e} x = 5\times(\log _{e}2 \times \log _{e}3)/(\log _{e}2 + \log _{e}3)\]

raise every term to the power of e

\[x = e^{(5\ln2\ln3)/(\ln2 + \ln3)}\]

Answer 2

this is a tough question becuase of the amount of manipulation needed