Question

((e^(x)-e^(-1))^2)/e^(2x) write expression so only one exponential function appears

Answer 1

Lets first simplify numerator only
can u expand ((e^(x)-e^(-1))^2)
using (a+b)^2 = a^2+2ab+b^2 formula?
what do u get ?

Answer 2

(e^x^2)-2+(e^(-x)^2)

Answer 3

sorry, the expression is ((e^(x)-e^(-x))^2)/e^(2x)

Answer 4

ok, then thats correct expansion of numerator.
but u can write
\((e^x)^2=e^{2x}\) ok ?
so your numerator will be
\(\huge e^{2x}-2+e^{-2x}\)
got this ?

Answer 5

ok

Answer 6

now u can distribute denominator's e^2x to each term of numerator .
\(\frac{e^{2x}}{e^{2x}}-\frac{2}{e^{2x}}+\frac{e^{-2x}}{e^{2x}}\)
ok?
and use
\(\huge \frac{e^{-2x}}{e^{2x}}=e^{-2x-2x}=?\)
so what u get ?

Answer 7

\[e ^{-4x}\]

Answer 8

so your expression will be
\(\huge 1-2e^{-2x}+e^{-4x} \)
got this?

Answer 9

ok

Answer 10

now again using (a+b)^2 = a^2+2ab+b^2 formula.,
taking a=1 and \(b= e^{-2x}\)
we can simplify that as
\(\huge (1-e^{-2x})^2\)
any doubts for this?

Answer 11

No

Answer 12

then thats your final answer.

Answer 13

Thank you!

Answer 14

welcome :)