# Math Help - logarithmic question

1. ## logarithmic question

having trouble with this one

solve for x using logs

2e^3x=4e^5x

thanks!

2. First, I would divide by 2 to produce:

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

Take the natural logarithm.

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

$3x = \ln(2) + 5x$

Subtract 5x.

$-2x = \ln(2)$

Divide by -2.

$x = -\frac{\ln(2)}{2}$

3. Helolo, aldric!

Another approach . . .

Solve for $x$ using logs: . $2e^{3x}\:=\:4e^{5x}$

We have: . $4e^{5x} \;=\;2e^{3x}$

Divide by $4e^{3x}\!:\;\;\frac{4e^{5x}}{4e^{3x}} \;=\;\frac{2e^{3x}}{4e^{3x}} \quad\Rightarrow\quad e^{2x} \;=\;\tfrac{1}{2}$

Take logs: . $\ln\left(e^{2x}\right) \;=\;\ln\left(\tfrac{1}{2}\right)$

. . . . . . . $2x\cdot\underbrace{\ln(e)}_{\text{This is 1}} \;=\;\ln\left(2^{-1}\right)$

. . . . . . . . . . . . $2x \;=\;-\ln(2)$

. . . . . . . . . . . . $x \;=\;-\tfrac{1}{2}\ln(2)$

4. i.e. $x=ln \left(\frac{1}{\sqrt{2}}\right)$.
At least, It has nice from. =)