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Let us say;
`y = log_ab`
Now remove log from right side.
`a^y = b`
Now take log on both side with respect to base b.
`log_b(a^y) = log_bb`
`ylog_ba = 1`
`y = 1/(log_ba)`
We know that `y = log_ab`
`log_ab = 1/(log_ba)` answer as required.
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