How do I solve #ln(1/2)^m = ln2#?

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Answer 1 · 2017-04-01T07:26:25

Two ways...

One is simply recognizing that by algebra, #2^(-1) = 1/2#. Hence, #m = -1# by inspection, i.e. #(2^(-1))^(-1) = 2 = (1/2)^(-1)#.

Another way is that anytime you see an exponent, try taking the #ln# of both sides.

#ln(1/2)^(m) = ln2#

Use the property that

#lna^b = blna#

Thus, we get:

#mln(1/2) = ln2#

#mln(2)^(-1) = ln2#

#-mln2 = ln2#

#=> m = -1#