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

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

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Answer 1 · 2016-04-02T02:36:29

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

Explanation:

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

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

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Explanation:

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