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e^x+e^-x=2.5

e^x+e^-x=2.5

So the way I have solved this is

e^x+1/e^-x-2.5=0
e^2x+1-2.5e^x=0

equating to y=e^x
y^2-2.5y+1=0
2y^2-5y+2=0
(2y-1)(y-2)
e^x=1/2 e^x=2

therefore x= ln0.5 or ln2

The textbook says the answer is +/-ln2.

Where did I go wrong?

Thanks

Reply 1

TenOfThem

Original post by elia

Where did I go wrong?

You didn't

You have the same answer

Reply 2

J10

You've got it correct. The answers are the same.

Reply 3

elia

Original post by J10

You've got it correct. The answers are the same.

Ah yes checked in the calculator, I always assume I'm wrong :/

Thanks guys!
/thread

Reply 4

TenOfThem

Original post by elia

Ah yes checked in the calculator, I always assume I'm wrong :/

Thanks guys!
/thread

You should not need to check on the calculator

ln(1/2) = - ln(2)

Rules of logs

(edited 10 years ago)

Reply 5

Mr M

Study Forum Helper

Original post by TenOfThem

You should not nee to check on the calculator

ln(1/2) = - ln(2)

Rules of logs

I'd have solved the equation

x = \cosh^{-1} \frac{5}{4}

(she can write the answer straight down from the formula book).

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e^x+e^-x=2.5

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