Question

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

Answer 1

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

Explanation:

`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)`.