SOLUTION: Please help me find all solutions (in non decimal form) of the equation: e^2x-8e^x+39/4=0 so far I have done: 4 (e^2x-8e^x+39/4)=0(4) 4e^2x-32e^x+39=0 I do not k

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Please help me find all solutions (in non decimal form) of the equation: e^2x-8e^x+39/4=0 so far I have done: 4 (e^2x-8e^x+39/4)=0(4) 4e^2x-32e^x+39=0 I do not k      Log On


   

Question 417918: Please help me find all solutions (in non decimal form) of the equation:
e^2x-8e^x+39/4=0
so far I have done:
4 (e^2x-8e^x+39/4)=0(4)
4e^2x-32e^x+39=0
I do not know how to procede onwards as a if solving for a quadratic equation because of 39
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Your first step is exactly what I would have done.
4e%5E2x-32e%5Ex%2B39=0
The next step will be to factor the left side. Once you realize that
  • since %28e%5Ex%29%5E2=+e%5E%282x%29 we can factor the left side just as you would factor 4q%5E2-32q%2B39; and
  • 39 is not prime, 39 = 13*3
you will see that just like
4q%5E2-32q%2B39
will factor into
(2q-13)(2q-3)
4e%5E2x-32e%5Ex%2B39
will factor into
%282e%5Ex-13%29%282e%5Ex-3%29
Now our equation is:
%282e%5Ex-13%29%282e%5Ex-3%29+=+0
From the Zero Product Property we know that one of these factors must be zero. So:
2e%5Ex-13+=+0 or 2e%5Ex-3+=+0
To solve these equations, with the variable in the exponent, we start by isolating the exponential terms. Adding 13 to each side of the first equation and 3 to each side of the second equation we get:
2e%5Ex+=+13 or 2e%5Ex+=+3
Dividing each side of each equation by 2 we get:
e%5Ex+=+13%2F2 or e%5Ex+=+3%2F2
Now we use logarithms. Although wa can use any base of logarithms, using a base of logarithm that matches the base of the exponent will result in a simpler expression. So we will use base e logarithms, aka ln:
ln%28e%5Ex%29+=+ln%2813%2F2%29 or ln%28e%5Ex%29+=+ln%283%2F2%29
On the left side of each equation we can use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponent of the argument (where the variable is) out in front (where we can then solve for the variable. It is this very property that is the reason we use logarithms to solve equations like these. Using this property we get:
x%2Aln%28e%29+=+ln%2813%2F2%29 or x%2Aln%28e%29+=+ln%283%2F2%29
By definition, ln(e) = 1 so these simplify to:
x+=+ln%2813%2F2%29 or x+=+ln%283%2F2%29
These are exact expressions for the solutions to your equation.