Question #645d2

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Answer 1 · 2017-07-06T20:05:36

I got: #x=14.09017#

I would use a property of logs as:
#logx+logy=log(xy)3 and write:

#log_5[(x-11)(x-6)]=2#

then use the definition of log to write:

#(x-11)(x-6)=5^2#

#(x-11)(x-6)=25#

rearrange:

#x^2-6x-11x+66-25=0#

#x^2-17x+41=0#

Solve using the Quadratic Formula:

#x_(1,2)=(17+-sqrt(289-164))/2=#

#x_1=14.09017#
#x_2=2.90903#

#x_2# is small and would make the arguments of the original logs negative.