SOLUTION: Solve the exponential equation algebraically. Show steps
2e^x=5-e^-x
Algebra.Com
Question 689684: Solve the exponential equation algebraically. Show steps
2e^x=5-e^-x
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
I am going to multiply each side by . You will better understand why if I first rewrite as :
Multiplying:
we get:
Since the exponent of is exactly twice the exponent of this equation is in what is called quadratic form. To see this better I am going to use a temporary variable. Let . Then . Substituting these in we get:
The equation is obviously quadratic. We first want one side to be zero. Subtracting the entire right side from each side we get:
Next we factor or use the Quadratic Formula. This will not factor so we must use the formula:
which simplifies as follows:
which is short for:
or
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?)
or
Now we solve these equations for x. (NOTE: Since 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 can never be zero or negative.)
We solve for x by finding the ln of each side:
or
On the left sides we use a property of logs, , to move the exponent out in front:
or
And since ln(e) = 1 these become:
or
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
to
to
etc.
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