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How do you solve #5^(2x-1) = 7^(2x)#?

1 Answer

Vinícius Ferraz

Jul 3, 2018

#x = frac{log 5}{2(log 5 - log 7)}#

Explanation:

#log 5^{2x - 1} = log 7^{2x}#

#(2x - 1) log 5 = 2x log 7#

#x (2 log 5 - 2 log 7) = 1 log 5#

#x = frac{log 5}{log 25 - log 49}#

#(25/49)^x = 5#

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