Precal

solve for x

7^x/2=5^1-x

what i have so far is:

(x/2)ln7=(1-x)ln5
im not sure where to go from there

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  1. [ln(7) / 2] x = ln(5) - [ln(5)] x

    x = ln(5) / {[ln(7) / 2] + ln(5)}

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  2. or, after some manipulation,

    1/(1+log5√7)

    is that simpler? who can say?

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