How do you solve #log_2[log_3(2x-5)]=log_4 16#?

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Answer 1 · 2017-05-22T22:01:57

#x=(3^4+5)/2#

#log_a a^b = b# then

#log_2[log_3(2x-5)]=log_4 16=2#

then

#log_3(2x-5)=2^2= 4# and

#2x-5 = 3^4# so finally

#x=(3^4+5)/2#

This is a feasible solution because

#2 ((3^4+5)/2)-5 gt 0#