solve for x
I just leraned this last spring and I finally get to use it
so what you do is like dis
remember we can solve ax^2+bx+c=0
get it into that form
sub u for e^x
2u^2-5u-3=0
solve for u
factor
(u-3)(2u+1)=0
set to zero
u-3=0
u=3
2u+1=0
2u=-1
u=-1/2
now
u=e^x so
e^x=3 and e^x=-1/2
e^x=-1/2 is not a valid solution because a^b will never be negative
e^x=3
take the ln of both sides
x=ln3
the solutiion is
x=ln3
