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log(base 4) 1/8=
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\[\text{Log}_4\left(\frac{1}{8}\right)\]
-1.5
I am not sure if there is a way to solve by hands
Change of base law.
@imran.
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\[\frac{\text{Log}_{10 }\frac{1}{8}}{\text{Log}_{10}4}\] Yeah , but you can't solve by hand
Yeah/
Ye of little faith.. \[Let\ k = log_4(\frac{1}{8})\] \[log_4(\frac{1}{8}) = k \iff 4^k = \frac{1}{8}\]\[\implies 4^k = 2^{-3}\]\[\implies (2^2)^k = 2^{-3}\]\[\implies 2^{2k} = 2^{-3}\]\[\implies 2k = -3\]\[\implies k = -\frac{3}{2}\]\[\implies log_4(\frac{1}{8}) = -\frac{3}{2} = -1.5\]
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