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

Answer 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.