How do you evaluate #3log_11 5+log_11 7#?

1 Answer
Alan P.
Apr 8, 2016

Simplified as #log_11(125/7)#
...beyond that a calculator is required to obtain: #~~1.202#

Explanation:

Some relations to remember:
#color(white)("XXX")p*log_b q= log_b q^p#

#color(white)("XXX")log_b s - log_b t = log_b s/t#

#3log_11 5color(white)("XXX")=log_11 5^3 = log_11 125#

Therefore
#3log_11 5 - log_11 7#
#color(white)("XXX")=log_11 125 - log_11 7#

#color(white)("XXX")=log_11 (125/7)#

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