How do you solve #2^(x+4) = 3^(2x+1)# ?

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Answer 1 · 2016-08-18T06:41:10

#x = (log 3 - 4 log 2)/(log 2 - 2 log 3) ~~ 1.112958972#

Taking logs of both sides, we have:

#(x+4)log 2 = log (2^(x+4)) = log (3^(2x+1)) = (2x+1)log(3)#

Multiplying out, this becomes:

#(log 2)x + 4 log 2 = (2log 3)x + log 3#

Hence:

#(log 2 - 2 log 3)x = log 3 - 4 log 2#

Divide both sides by #(log 2 - 2 log 3)# to get:

#x = (log 3 - 4 log 2)/(log 2 - 2 log 3) ~~ 1.112958972#