# Variable in exponent

• Apr 29th 2013, 03:05 AM
Unreal
Variable in exponent
Hi!

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

I have used $\displaystyle \log_3$ but this is where it leads:

$\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

$\displaystyle 2x \log_3 e = x - 4$

$\displaystyle 2x \log_3 e + 4 = x$

$\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac12$
• Apr 29th 2013, 03:28 AM
topsquark
Re: Variable in exponent
Quote:

Originally Posted by Unreal
Hi!

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

I have used $\displaystyle \log_3$ but this is where it leads:

$\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

$\displaystyle 2x \log_3 e = x - 4$

$\displaystyle 2x \log_3 e + 4 = x$

$\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac12$

What happened in that last line?

$\displaystyle 2x \log_3 e + 4 = x$

$\displaystyle 2x \log_3 e - x = -4$

Now factor the x on the LHS and solve.

-Dan
• Apr 29th 2013, 07:39 AM
HallsofIvy
Re: Variable in exponent
Quote:

Originally Posted by Unreal
Hi!

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

I have used $\displaystyle \log_3$ but this is where it leads:

$\displaystyle \log_3 e^{2x} = \log_3 3^{x-4}$

$\displaystyle 2x \log_3 e = x - 4$

$\displaystyle 2x \log_3 e + 4 = x$

This is correct up to here although I think it would have been better to take the logarithm with respect to e, not 3, getting
2x= (x+ 3) ln(3) just because you can easily find a numerical value for log(3). But, as I say, what you have done is correct to this point.

Quote:

$\displaystyle \frac{\log_3 e + 4}{2} = \frac{x}{2x} = \frac{1}{2}$
But here you have gone wrong. Dividing the left side by "2x" leaves $\displaystyle log_3(e)+ \frac{2}{x}= \frac{1}{2}$

Subtract $\displaystyle log_3(e)$ from both sides to get $\displaystyle \frac{2}{x}= \frac{1}{2}- log_3(x)$ and then, since "x" is in the denominator on the left, take the reciprocal of both sides:
$\displaystyle \frac{x}{2}= \frac{1}{\frac{1}{2}- log_2(x)}$

$\displaystyle x= \frac{2}{\frac{1}{2}- log_3(e)}$