Math

How do I solve e^(2x)+10e^(x)-75=0 and 3^(x+2)=7^(x-5)?

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  1. For e^(2x)+10e^(x)-75=0
    Let
    y=e^x
    then
    y² = e^(2x)
    Substitute in above equation,
    y²+10y-75=0
    Factor and solve for y.
    Note that the domain of y (=e^x) is (0,∞), so any negative root for the quadratic equation must be rejected.

    For 3^(x+2)=7^(x-5)
    take logarithm on both sides, and apply the exponent rule to get
    (x+2)*ln 3 = (x-5)*ln 7
    Solve for x. I get x=14 approx.

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