How do you evaluate #log 0.5915#?

1 Answer
Alan P.
Sep 2, 2015

#log_10 0.5915 = -0.22805#

Explanation:

Possibly not the kind of answer you were hoping for, but the only reasonable way to do this is to use a calculator (that has a log function).

You are looking for a value, #x# such that
#color(white)("XXXX")10^x = 0.5915#

Since #10^0=1 and 10^(-1)=0.1#
and since #1 > 0.5915 > 0.1#
#rarr color(white)("XXXX") 0 > x > -1#

We could try different values between #0 and -1# as the exponent, but without a calculator (or similar technology) such evaluations are too time consuming to be reasonable.

Related questions
See all questions in Logarithm-- Inverse of an Exponential Function
Impact of this question
1207 views around the world
You can reuse this answer
Creative Commons License