Question

How do you solve `log_2x=log_5 3`?

Answer 1

`x=1.605`

Explanation:

`log_2x=log_53` can be simplified using `log_ba=loga/logb`. Hence it is

`logx/log2=log3/log5`

or `logx=log3/log5xxlog2`

or `logx=0.4771/0.6990xx0.3010`

Hence `x=10^(0.4771/0.6990xx0.3010)=1.605`