SOLUTION: e^x-e^-x=10

Question 719905: e^x-e^-x=10
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

Rewriting as will help us see what to do:

First, we'll eliminate the fraction by multiplying both sides by :

which simplifies to:

Next we'll get a zero on the right side by subtracting :

This does not factor, unfortunately.

This next part is probably the hardest part. The exponent of is exactly twice the exponent of . This makes this equation an equation in what is called "quadratic form". Quadratic form equations can be solved in much the same way as regular quadratic equations. Using a temprary variable can help you see the "quadratic-ness" of our equation. Set it equal to the base and the smaller exponent:
Let . Then . Substituting these into our equation we get:

This is clearly an quadratic equation. As noted earlier, this will not factor. But we can use the quadratic formula:

Simplifying...

which is short for:
or

Of course we are not interested in solutions for q. We are interested in solutions for x. So now we substitute back for the q:
or
We have a little more work to do to solve for x. Since the second equation says that is a negative number. But it is impossible for e to any power to be a negative number. So that equation is impossible. There will be no solutions for x from that equation.

So our only solution(s) will come from the first equation. Finding the ln of each side:

On the left side we use a property of logs, , to move the exponent out in front:

Since ln(e) = 1 the left side becomes:

This is an exact expression for the solution to your equation. If you want/need a decimal approximation, get out your calculator.

P.S. After you have done a few of these quadratic form equations, you will not need a temporary variable. You will start to see how to go directly from:

etc.
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