Question

I've recently taken a math quiz, and there was one problem that wouldn't get out of my head:

e^(2x+1)=6*e^(x+3) Solve for x.

My answer was: x=ln(6)+2
I now know that's the wrong answer and a calculator states it is x=2.

Could anyone perhaps tell me how to solve for x? In a step-by-step process; if another problem comes up in the future, I'd like to think back and remember all the steps.

Answer 1

You are actually correct?
http://www.wolframalpha.com/input/?i=e%5E%282x%2B1%29%3D6*e%5E%28x%2B3%29

Answer 2

divide both sides by
\[e^{x+3}\]

Answer 3

get
\[\frac{e^{2x+1}}{e^{x+3}}=6\]

Answer 4

subtract the exponents , get
\[e^{x-2}=6\] making
\[x-2=\ln(6)\] or
\[x=\ln(6)+2\]

Answer 5

yeap.... got the same here as well

Answer 6

if it was
\[e^{2x+1}=e^{x+3}\] THEN it would be \(x=2\)

Answer 7

I'm reading \(\Large \bf e^{2x+1}=6\cdot e^{x+3}\) if so, then ln(6)+2 is right

Answer 8

Yay!, Mathway Calculator stated x=2, and my math quiz was 3 questions, I thought i was so screwed.

Answer 9

Thank you, everyone.

Answer 10

yw

Answer 11

Here's my way, if you are interesting...:

I substitute 6 to \(\large e^{\ln6}\) so we have:
\(\large e^{x-2}=e^{\ln6}e^{x+3} \\~\\ \large \large e^{x-2}=e^{x+3+\ln6} \\~\\x-2=x+3+\ln6\)

You can do the rest from here

Answer 12

Oops i typed x instead of 2x in left side of equation... sorry

Answer 13

actually 2x+1, not x-2

haha sorry about that... but you get the idea; replace 6 to \(e^{ln6}\)

Answer 14

Thank you! I didn't realize there were so many ways to do it.
On the quiz i started off by using the inverse ln:
ln(e^(2x+1))=ln(6)*ln(e^(x+3))
=ln(e^2x+1))=ln(6+e^(x+3))
=(2x+1)=ln(6)+(x+3)

and had continued from there.