SOLUTION: Solve the exponential equation algebraically. Show steps 2e^x=5-e^-x

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Question 689684: Solve the exponential equation algebraically. Show steps
2e^x=5-e^-x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2e%5Ex=5-e%5E%28-x%29
I am going to multiply each side by e%5Ex. You will better understand why if I first rewrite e%5E%28-x%29 as 1%2Fe%5Ex:
2e%5Ex=5-1%2Fe%5Ex
Multiplying:
e%5Ex%2A%282e%5Ex%29=e%5Ex%2A%285-1%2Fe%5Ex%29
we get:
2e%5E%282x%29=5e%5Ex+-1
Since the exponent of e%5E%282x%29 is exactly twice the exponent of e%5Ex this equation is in what is called quadratic form. To see this better I am going to use a temporary variable. Let q+=+e%5Ex. Then q%5E2+=+%28e%5Ex%29%5E2+=+e%5E%282x%29. Substituting these in we get:
2q%5E2+=+5q-1
The equation is obviously quadratic. We first want one side to be zero. Subtracting the entire right side from each side we get:
2q%5E2-5q+%2B1=0
Next we factor or use the Quadratic Formula. This will not factor so we must use the formula:
q+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%282%29%281%29%29%29%2F2%282%29
which simplifies as follows:
q+=+%28-%28-5%29+%2B-+sqrt%2825-4%282%29%281%29%29%29%2F2%282%29
q+=+%28-%28-5%29+%2B-+sqrt%2825-8%29%29%2F2%282%29
q+=+%28-%28-5%29+%2B-+sqrt%2817%29%29%2F2%282%29
q+=+%285+%2B-+sqrt%2817%29%29%2F4
which is short for:
q+=+%285+%2B+sqrt%2817%29%29%2F4 or q+=+%285+-+sqrt%2817%29%29%2F4
We now have solutions for q. But we are looking for solutions for x. Now we substitute back in for q. (It was temporary, remember?)
e%5Ex+=+%285+%2B+sqrt%2817%29%29%2F4 or e%5Ex+=+%285+-+sqrt%2817%29%29%2F4
Now we solve these equations for x. (NOTE: Since sqrt%2817%29 is less than 5, the right side of both equations will be positive. If we had gotten zero or a negative for the right side of either equation there would be no solution for that equation since e%5Ex can never be zero or negative.)

We solve for x by finding the ln of each side:
ln%28e%5Ex%29+=+ln%28%285+%2B+sqrt%2817%29%29%2F4%29 or ln%28e%5Ex%29+=+ln%28%285+-+sqrt%2817%29%29%2F4%29
On the left sides 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%28%285+%2B+sqrt%2817%29%29%2F4%29 or x%2Aln%28e%29+=+ln%28%285+-+sqrt%2817%29%29%2F4%29
And since ln(e) = 1 these become:
x+=+ln%28%285+%2B+sqrt%2817%29%29%2F4%29 or x+=+ln%28%285+-+sqrt%2817%29%29%2F4%29
These are exact expressions for the solutions to your equation. If you want/need decimal approximations get out your calculator.

P.S. Once you have done a few of these quadratic form equations, you will not need to use a temporary variable. You will see how to go from
2e%5E%282x%29=5e%5Ex+-1
to
2e%5E%282x%29-5e%5Ex+%2B1=0
to
e%5Ex+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%282%29%281%29%29%29%2F2%282%29
etc.