Question #c11ad

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Answer 1 · 2016-12-26T19:53:20

see below

concepts applied

if referring to the log as base 10 log, #log_10x#:
#(\log_10(225))/(\log_10(25))#

then
numerator: #\log_10(15^2)\rArr2\log_10(15)#
denominator: #\log_10(5^2)\rArr2\log_10(5)#

becomes
#(2\log_10(25))/(2\log_10(5))#
the 2's cancel, which leaves you with #(\log_10(25))/(\log_10(5))#

simplify and you get your answer.

if referring to the log as natural log, #log_ex# or #lnx#:
#(\log_e(225))/(\log_e(25))\rArr(\ln(225))/(\ln(25))#

then
numerator: #\ln(15^2)\rArr2\ln(15)#
denominator: #\ln(5^2)\rArr2\ln(5)#

becomes
#(2\ln(25))/(2\ln(5))#
the 2's cancel, which leaves you with #(\ln(25))/(\ln(5))#

simplify and you get your answer.