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

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Question 719905: e^x-e^-x=10
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
e%5Ex-e%5E%28-x%29=10
Rewriting e%5E%28-x%29 as 1%2Fe%5Ex will help us see what to do:
e%5Ex-1%2Fe%5Ex=10
First, we'll eliminate the fraction by multiplying both sides by e%5Ex:
e%5Ex%2Ae%5Ex-e%5Ex%2A%281%2Fe%5Ex%29+=+e%5Ex%2A10
which simplifies to:
e%5E%282x%29-1+=+10e%5Ex
Next we'll get a zero on the right side by subtracting 10e%5Ex:
e%5E%282x%29-10e%5Ex-1+=+0
This does not factor, unfortunately.

This next part is probably the hardest part. The exponent of e%5E%282x%29 is exactly twice the exponent of 10e%5Ex. 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 q+=+e%5Ex. Then q%5E2+=+%28e%5Ex%29%5E2+=+e%5E%282x%29. Substituting these into our equation we get:
q%5E2-10q-1=0
This is clearly an quadratic equation. As noted earlier, this will not factor. But we can use the quadratic formula:
q+=+%28-%28-10%29+%2B-+sqrt%28%28-10%29%5E2-4%281%29%28-1%29%29%29%2F2%281%29
Simplifying...
q+=+%28-%28-10%29+%2B-+sqrt%28100-4%281%29%28-1%29%29%29%2F2%281%29
q+=+%28-%28-10%29+%2B-+sqrt%28100%2B4%29%29%2F2%281%29
q+=+%28-%28-10%29+%2B-+sqrt%28104%29%29%2F2%281%29
q+=+%2810+%2B-+sqrt%28104%29%29%2F2
q+=+%2810+%2B-+sqrt%284%2A26%29%29%2F2
q+=+%2810+%2B-+sqrt%284%29%2Asqrt%2826%29%29%2F2
q+=+%2810+%2B-+2%2Asqrt%2826%29%29%2F2
q+=+%282%285+%2B-+sqrt%2826%29%29%29%2F2
q+=+%28cross%282%29%285+%2B-+sqrt%2826%29%29%29%2Fcross%282%29
q+=+5+%2B-+sqrt%2826%29
which is short for:
q+=+5+%2B+sqrt%2826%29 or q+=+5+-+sqrt%2826%29

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:
e%5Ex+=+5+%2B+sqrt%2826%29 or e%5Ex+=+5+-+sqrt%2826%29
We have a little more work to do to solve for x. Since sqrt%2826%29+%3E+5 the second equation says that e%5Ex 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:
ln%28e%5Ex%29+=+ln%285+%2B+sqrt%2826%29%29
On the left side we use a property of logs, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent out in front:
x%2Aln%28e%29+=+ln%285+%2B+sqrt%2826%29%29
Since ln(e) = 1 the left side becomes:
x+=+ln%285+%2B+sqrt%2826%29%29
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:
e%5E%282x%29-10e%5Ex-1+=+0
to
e%5Ex+=+%28-%28-10%29+%2B-+sqrt%28%28-10%29%5E2-4%281%29%28-1%29%29%29%2F2%281%29
etc.