# natural exponential function with two solutions

• Jan 25th 2007, 07:44 PM
thedoge
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
• Jan 25th 2007, 09:41 PM
earboth
Quote:

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:

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

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

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

Now re-substitute:

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

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

EB
• Jan 26th 2007, 03:47 AM
topsquark
Quote:

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)]

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