1. ## Exponential equations.

The goal of the problem is to solve for x using logs and the given equation is this:

2e^(3x) = 4e^(5x)

Any help on how to solve this?

2. Originally Posted by Rumor
The goal of the problem is to solve for x using logs and the given equation is this:

2e^(3x) = 4e^(5x)

Any help on how to solve this?
I'll start you off;

$\displaystyle 2e^{3x}=4e^{5x}$

$\displaystyle e^{3x}=2e^{5x}$

$\displaystyle ln(e^{3x})=ln(2e^{5x})$

$\displaystyle 3x=ln(2)+ln(e^{5x})$

Can you take it from here?

Keep in mind the handy rules $\displaystyle log_{a}a=1$, and $\displaystyle log_{a}b^c=c \cdot log_{a}b$

3. Originally Posted by Kasper
I'll start you off;

$\displaystyle 2e^{3x}=4e^{5x}$

$\displaystyle e^{3x}=2e^{5x}$

$\displaystyle ln(e^{3x})=ln(2e^{5x})$

$\displaystyle 3x=ln(2)+ln(e^{5x})$

Can you take it from here?

Keep in mind the handy rules $\displaystyle log_{a}a=1$, and $\displaystyle log_{a}b^c=c \cdot log_{a}b$
Yes, I was able to solve it from there.

Thank you! :]