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

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Answer 1 · 2016-09-24T18:59:05

#x = 16.4695#

As the bases are different, we cannot just compare them.

The variables are in the exponents, so logs are called for.

Log both sides:

#log2^(x+3) = log3^(x-4)" "larr# use the log power law

#(x+3)log2 = (x-4)log3 " "larr# move the log terms to one side

#(x+3)/(x-4) = (log3)/(log2) = 1.58496#

#(x+3)/(x-4) = 1.58496" "larr# cross-mulitply

#x+3 = 1.58496(x-4)#

#x +3 = 1.58496x -6.33985" "larr# re-arrange the terms

#3+6.33985 = 1.58496x -x#

#9.633985 = 0.58496x#

#9.633985/0.58496 = x#

#x = 16.4695#