How do you solve #7^x=5^(x-4)#?

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Answer 1 · 2015-08-22T08:20:53

Take #log# of both sides and use properties of #log# of exponents to find:

#x = (-4log(5))/(log(7) - log(5)) ~~ -19.133#

Take #log# of both sides to get:

#x log(7) = log(7^x) = log(5^(x-4)) = (x - 4) log(5)#

#= x log(5) - 4 log(5)#

Subtract #x log(5)# from both sides to get:

#x (log(7) - log(5)) = -4log(5)#

Divide both sides by #(log(7) - log(5))# to get:

#x = (-4log(5))/(log(7) - log(5)) ~~ -19.133#