Must find x

Ae^Bx = e^Cx

Where A,B,C are constants

My work:

lnA + Bx = Cx

lnA / (C - B) = x

I am not confident in this answer

Results 1 to 4 of 4

- Apr 8th 2010, 05:09 PM #1

- Joined
- Apr 2010
- Posts
- 2

- Apr 8th 2010, 05:16 PM #2
I get

$\displaystyle Ae^{Bx} = e^{Cx}$

$\displaystyle \ln (Ae^{Bx}) = \ln (e^{Cx})$

$\displaystyle \ln (A) +\ln (e^{Bx})= \ln (e^{Cx})$

$\displaystyle \ln (A) +Bx= Cx$

$\displaystyle \ln (A) = Cx-Bx$

$\displaystyle \ln (A) = x(C-B)$

$\displaystyle \frac{\ln (A)}{C-B} = x$

$\displaystyle x=\frac{\ln (A)}{C-B} $

Same as you, with are either both right or both wrong!

- Apr 8th 2010, 05:34 PM #3
Let me try

$\displaystyle Ae^{Bx} = e^{Cx}$

$\displaystyle \ln{(Ae^{Bx})} = \ln{(e^{Cx})}$

$\displaystyle \ln{(A)} + \ln{(e^{Bx})} = \ln{(e^{Cx})}$

$\displaystyle \ln{(A)} + Bx = Cx$

$\displaystyle \ln{(A)} = Cx - Bx$

$\displaystyle \ln{(A)} = x(C - B)$

$\displaystyle \frac{\ln{(A)}}{C - B} = x$

$\displaystyle x = \frac{\ln{(A)}}{C - B}$

Wolfram says so too. Looks like this is the answer

**EDIT**: Strange, I did exactly the same steps as Pickslides and in the same order ... but I didn't look at all.

- Apr 8th 2010, 06:08 PM #4

- Joined
- Apr 2010
- Posts
- 2