1. ## Solve for X

Solve for X

X: E^(2x)-(5E^x)-6=0

You don't have to solve it for me, but if I could get a hint it would help. Thanks ahead of time.

2. ## Re: Solve for X

take
$e^x=y$

then $y^2-5y-6=0$ (a normal quadric equation)

3. ## Re: Solve for X

So I substitute and use the quadratic equation. And I get say Y=6 or Y=-1. Then I say E^x=6 or E^x=-1. Do I now say, Log e (6)=x, or log e (-1)=x

And because log e (-1)=x, I can say that E must equal in this case=-1, and x must be an odd integer?

So, X=all odd integers, or X=log e (6)? Depending on which solution for E you use? Is this correct, and if so how do I properly write this answer?

4. ## Re: Solve for X

Originally Posted by StudentMCCS
$\displaystyle \ln{x}$ is only defined for $\displaystyle x > 0$...