# Thread: Need help with exponential equation

1. ## Need help with exponential equation

Hello,

I have a job interview tomorrow, where I will be given a maths test. It's been several years since I did maths last and I'm quite rusty. I'm having difficulty with the following problem in the practice paper:

$e ^ (2x) + 2e^x = 8$

My attempt went as follows:

ln e ^ (2x) + ln 2 e ^ x = ln 8

ln 2 e ^ (3x) = ln 8

2 e ^ (3x) = 8

e ^ (3x) = 4

3x = ln 4

x = (ln 4)/3

However when I substitute this back in, it doesn't seem to work. Where am I going wrong?

2. My attempt went as follows:

ln e ^ (2x) + ln 2 e ^ x = ln 8

No, that is not correct.
ln(a +b) is not ln(a) +ln(b).
Rather, ln(a*b) = ln(a) +ln(b)

By the way, your equation do not need logs yet.
If you look closely, it is a quadratic equation in e^x

e^(2x) +2e^x = 8
Or, that is,
(e^x)^2 +2e^x -8 = 0

It is like having
y^2 +2y -8 = 0
if y = e^x

So, factoring that,
(y +4)(y-2) = 0

y +4 = 0
y = -4
Or, back to the e^x for y,
e^x = -4
Take the ln of both sides,
ln(e^x) = ln(-4)
x*ln(e) = ln(-4)
x = ln(-4)
But there are no logs of negative numbers (no real number logarithmns for negative numbers), so reject e^x = -4

y -2 = 0
Or,
e^x -2 = 0
e^x = 2
x = ln(2) = 0.693147181 -----------------answer.

3. hello, here's an answer for you

e^2x + 2e^x = 8

e^2x + 2e^x -8 = 0 (treat it like a normal quadratic)

(e^x +4)(e^x-2) = 0

therefore
e^x +4 = 0 or e^x -2 = 0

e^x-2 = 0
ln e^x = ln 2
x = ln 2

as the x in e^x +4 = 0 turns out to be ln(-4) which can't be, we ingore this answer meaning that it has to be the above answer.

i think your problem was that when you loged the equation, you gave both e's a seprate ln when they should have been made into ln( e^2x + 2e^x) which would be impossible to split up into seprete lns.

hope that helps, (good luck in the interview)

4. Thank you both; thats perfect

I'm rustier than I thought