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

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Answer 1 · 2016-04-02T02:36:29

$x = log 5/(log 5 +(1/2)log 7)$ = 0.623235, nearly.

$log (a^n)=nlog a$
equating the logarithms,
$(x/2) log 7 = (1-x) log 5$
Solve for the inverse relation $x = log 5/(log 5 +(1/2)log 7)$ .

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