Evaluate log2^(1÷8)?

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Answer 1 · 2018-05-12T05:08:32

-.60205...

#log2^(1/8)# intends that it is a common log, making it the same as #log_(10)2^(1/8)#

First, let's simplify the 2.

#2^(1/8)#=#1/4#

So, #log_(10)(1/4)#

This can also be written in exponential form, yielding:

#10^x#=#1/4#

Either way, the answer comes out to be roughly -.60205...