How do you evaluate #log_5(1/2)#?

1 Answer
EZ as pi
May 2, 2016

By calculator -0.431 (3 dec places)

Explanation:

Let #log_5 (1/2) = x#

Change from log form to index from: #5^x = 1/2#

Now, because the variable is the index, and #1/2# is not an exact power of 5, we need to use logs.

Power law: #x log5# = log 0.5)

#x = (log0.5)/(log5)#

From a calculator #x# = -0.431

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