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^(x5)?

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+10y75=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^(x5)
take logarithm on both sides, and apply the exponent rule to get
(x+2)*ln 3 = (x5)*ln 7
Solve for x. I get x=14 approx.
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