Question

question below:

Answer 1

using the substitution \[y = 2e ^{x}\]
solve:
\[e ^{x} = 7 - 12e ^{-x}\]

Answer 2

multiply both sides by e^x
u get
\(e^{(2x)}=7e^x-12 \\
now \quad e^x=y/2 \\
e^{2x}=y^2/4 \\ y^2/4-7(y/2)+12=0\)
can u solve this quadratic in y ??

Answer 3

u get \(y^2-14x+48=0\)
which is easily factorable.

Answer 4

hold on.. in the first step.. why would you multiply both sides by e^x?

Answer 5

I am SO sorry. Typo! the substitution had to be y= e^x.
NOT 2e^x

Answer 6

i would multiply both sides by e^x to get the e^{2x} term for substitution
\(e^{(2x)}=7e^x-12 \\
now \quad e^x=y \\
e^{2x}=y^2 \\ y^2-7(y)+12=0\)

Answer 7

still easily factorable........

Answer 8

Okay yes! I get it! Thank you so much for your help! :)