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

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

How do you solve 7^(x-2)=5^x7^(x-2)=5^x?