Posted by **Anonymous** on Sunday, December 20, 2009 at 8:15pm.

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

- Pre Calc -
**MathMate**, Sunday, December 20, 2009 at 9:29pm
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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