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

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Answer 1 · 2016-07-01T08:36:08

#x=2 /(1 - log_7 5)#

#7^(x-2)=5^x#

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

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

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

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

specifically choosing base 7 for the log

#x=2 /(1 - log_7 5)#