excaly Posts: 3, Reputation: 1 New Member #1 May 12, 2010, 06:06 PM
Quadratic: Solve (e^x - e^-x)/2 = 2 (sinhx = 2)
I know x = 1.443 but I need to solve it now using algebra, not my calculator.

Pauls Online Notes : Algebra - Solving Exponential Equations
Algebra II: Exponential and Logarithmic Functions - Math for Morons Like Us

These are 2 of the sites I've been on, I've also gone through my textboox but I can't find any information for when you have two e's and two uses of x in the equation.

This is as far as I got.
(e^x - e^-x)/2 = 2
e^x - e^-x = 4
ln4= lne^x - lne^-x

Thanks in advance, I'm really struggling here.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 May 13, 2010, 04:49 AM

Are you familiar with Newton's Method?
 excaly Posts: 3, Reputation: 1 New Member #3 May 13, 2010, 06:46 AM
No I am not
 ebaines Posts: 12,130, Reputation: 1307 Expert #4 May 13, 2010, 07:13 AM

Newton's method is an approximation technique - very effective at getting an estimate of the answer to as many decimals of accuracy as you like. However, there is a way to get an exact solution here. Multiply through by e^x and rearrange, and you get:

$
e^{2x} - 4 e^x -1 = 0
$

Now substitute $u = e^x$:

$
u^2 - 4u -1 = 0
$

Solve using the quadratic equation, but remember that only the positive solution of u is valid. Then convert back to x using $x = \ln(u)$

You will find that $x = \ln(2 + \sqrt{5})$

One other point - in your original attempt you wrote this:

$
e^x - e^{-x} = 4 \\
\ln(4) = \ln(e^x) - \ln(e^{-x})
$

But that is incorrect. If you take the log of the left hand side, you must also take the log of the right hand side as a whole:

$
\ln(4) = \ln(e^x - e^{-x})
$

But log(a-b) does NOT equal log(a) - log(b). So your last step is invalid.
 excaly Posts: 3, Reputation: 1 New Member #5 May 13, 2010, 07:18 AM

Wow thank you very much ebaines and galactus.

Ill spend a while on here tomorrow and see if I can help out any other members :)

Appreciated.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #6 May 13, 2010, 11:36 AM

Ebaine's solution I think is best, but here is the Newton's method I mentioned. It comes in handy when the equations are hard to solve algebraically. This one can be though.

$x_{n+1}=x_{n}-\frac{f(x)}{f'(x)}$

The derivative of $e^{x}-e^{-x}-4=0$ is 2cosh(x)

So, we have:

$x-\frac{2sinh(x)-4}{2cosh(x)}$

Use an initial guess of say, 1.5

Plugging that in the above we get

1.44504381624

Now, iterate by plugging this into the formula. Then, we get:

1.44363636201

There. Close enough.

That is the solution.

We could also have tried another guess of , say, 1. It would converge to the same solution, though not quite as fast.

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