How do you evaluate 3log_11 5+log_11 73log115+log117?

1 Answer
Apr 8, 2016

Simplified as log_11(125/7)log11(1257)
...beyond that a calculator is required to obtain: ~~1.2021.202

Explanation:

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

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

3log_11 5color(white)("XXX")=log_11 5^3 = log_11 1253log115XXX=log1153=log11125

Therefore
3log_11 5 - log_11 73log115log117
color(white)("XXX")=log_11 125 - log_11 7XXX=log11125log117

color(white)("XXX")=log_11 (125/7)XXX=log11(1257)