# Thread: natural exponential function with two solutions

1. ## natural exponential function with two solutions

e^(2x)-6e^x+5=0 has two solutions, what are they?

I tried this:
1. moved +5 to the other side so there wouldnt be a ln 0 for the next step
2. applied ln to both sides [2x-6x=ln(-5)]
3. combined like terms = -4x=ln(-5)
4. isolated x = x=ln(-5)/-4

obviously this is wrong

the only thing i can think of that would yield two answers is the quadratic formula, but i don't know if thats the right route or how to set that up

edit still stuck here. any tips on where to start?

edit2: Ok. I just looked at this problem and realized that e^0 would work for one answer. I assume the other I can work out now.

edit3: still stuck. tried everything i know

2. Originally Posted by thedoge
e^(2x)-6e^x+5=0 has two solutions, what are they?
...
Hello,

to solve this equation you have to use substitution:

$\displaystyle e^{2x}-6e^x+5=0$
$\displaystyle (e^{x})^2-6e^x+5=0$

Now set $\displaystyle e^x=y$. Your equation becomes now:

$\displaystyle (y)^2-6y+5=0$ which have the solutions y = 1 or y = 5

Now re-substitute:

$\displaystyle e^x=1 \Longrightarrow x = 0$ or

$\displaystyle e^x=5 \Longrightarrow x = \ln(5)$

EB

3. Originally Posted by thedoge
e^(2x)-6e^x+5=0 has two solutions, what are they?

I tried this:
1. moved +5 to the other side so there wouldnt be a ln 0 for the next step
2. applied ln to both sides [2x-6x=ln(-5)]
$\displaystyle e^{2x} - 6e^x + 5=0$

$\displaystyle e^{2x} - 6e^x = -5$

$\displaystyle ln \left ( e^{2x} - 6e^x \right ) = ln(-5)$

1) ln(-5) doesn't exist. The domain for the ln function is $\displaystyle (0, \infty)$.
2) $\displaystyle ln \left ( e^{2x} - 6e^x \right ) \neq ln(2x - 6x)$. You are trying to mix a valid rule with a mistaken one here. Yes, $\displaystyle ln \left ( e^{2x} \right ) = 2x$, but $\displaystyle ln(a^b + c)$ cannot be simplified in general.