# calc II, but stuck on algebra

• Sep 20th 2005, 05:53 PM
msm1593
a problem for calc II, but stuck on algebra
alright, so in this problem i need to solve for A.

y = 4e^x + e^-x
y = Acosh(x + ln2)

now you might already know cosh(x) = (e^x + e^-x) / 2

and with a little fooling around i found that e ^ ln2 = 2
and that cosh(ln(2)) = 1.25 (?)

so i think i have most of the pieces figured out but i cannot put them together. after the substitution where should i go with this?
• Sep 21st 2005, 05:02 AM
ticbol
Let me use the datas/informations as posted. If they are wrongly posted, then my answer will be wrong also.

y = 4e^x + e^-x -----------(1)
y = A*cosh(x + ln2) --------------(2)

-----------------
cosh(ln(2))
= [e^ln(2) +e^(-ln(2))]/2 ----(i)

Let e^ln(2) = N
Take the ln of both sides,
ln(2)*ln(e) = ln(N)
ln(2) = ln(N)
2 = N
So, e^ln(2) = 2 -----***

Let e^[-ln(2)] = M
1/[e^ln(2)] = M
1/[2] = M
So, e^[-ln(2)] = 1/2 -----***

Substitute those into the cosh(ln(2)),
cosh(ln(2))
= [e^ln(2) +e^(-ln(2))]/2 ----(i)
= [2 +1/2]/2
= 5/4 or 1.25 ---------***

-----------------------------
sinh(ln(2))
= [e^ln(2) -e^(-ln(2))]/2 ----(ii)
= [2 -1/2]/2
= 3/4 or 0.75 ---------***

---------------------------------------------
y = A*cosh(x + ln2) --------------(2)
y = A*{cosh(x)cosh[ln(2)] +sinh(x)sinh[ln(2)]}
y = A*{[(e^x +e^-x)/2](5/4) +[(e^x -e^-x)/2](3/4)}
y = (A/8){(e^x +e^-x)(5) +(e^x -e^-x)(3)}
y = (A/8){5e^x +5e^-x +3e^x -3e^-x}
y = (A/8){8e^x +2e^-x}
y = (A/4){4e^x +e^-x} ----------(2a)

y from (2a) = y from (1),
(A/4){4e^x +e^-x} = 4e^x +e^-x
Divide both sides by (4e^x +e^-x),
A/4 = 1
A = 1*4 = 4 --------------------------answer.