How do you evaluate #g(x)=log_bx# for #x=b^-3#?

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How do you evaluate #g(x)=log_bx# for #x=b^-3#?

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Answer 1 · 2017-01-18T18:13:48

#g(b^-3) = -3#

Explanation:

In general, #log_a a^n = n#, because #log_a b# is the power #a# must be raised to, to get #b#. So, in our case, #-3# is the power #b# needs to be raised, in order to get #b^-3#. #log_b b^-3 = -3#.

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How do you evaluate #g(x)=log_bx# for #x=b^-3#?

1 Answer

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