How do you evaluate ${log}_{4} 4^{- 0.38}$ $log_4 4^-0.38$ ?

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Answer 1 · 2016-11-30T22:04:48

I got $- 0.38$ $-0.38$

We can take the exponent in front of the log and write:

$- 0.38 {log}_{4} (4) =$ $-0.38log_4(4)=$

but ${log}_{4} (4) = 1$ $log_4(4)=1$ (from the definition of log) so you get:

$- 0.38 \cdot 1 = - 0.38$ $-0.38*1=-0.38$

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