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.

Are you familiar with Newton's Method?

No I am not

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:



Now substitute :



Solve using the quadratic equation, but remember that only the positive solution of u is valid. Then convert back to x using

You will find that

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



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:



But log(a-b) does NOT equal log(a) - log(b). So your last step is invalid.

Wow thank you very much ebaines and galactus.

Very very helpful.

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

Appreciated.

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.



The derivative of is 2cosh(x)

So, we have:



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.