Logarithm + Nepper's number Exercise

[Velo]

" Solve the equation:

2e^(-x) = 3e^(0.1x) "

I've been fiddling around with this and I have no idea what I'm supposed to do. I know the final answer should be something like:
x = [(2/3)log e] / 1,1

The only step I've managed to do was:
(2/3) x e^(-x) = e^(0.1x)

But after that, I don't know how to get the x out of the exponential.. Help would be appreciated :')

 

[MarkFL]

We are given to solve:

\(\displaystyle 2e^{-x}=3e^{0.1x}\)

My first step would be to multiply though by $e^x\ne0$ to get:

\(\displaystyle 2=3e^{1.1x}\)

Now divide though by 3:

\(\displaystyle \frac{2}{3}=e^{1.1x}\)

Next, what do you get when converting from exponential to logarithmic form?

 

[Velo]

Hm...

1.1x = log e ^ (2/3) <=>
x = [log e ^ (2/3)]/1.1 <=>
x = [(2/3) log e]/1.1

Which was the solution Is that right? Also, sorry, I'm having trouble using the latex thingy ><

 

[MarkFL]

Hm...

1.1x = log e ^ (2/3) <=>
x = [log e ^ (2/3)]/1.1 <=>
x = [(2/3) log e]/1.1

Which was the solution Is that right? Also, sorry, I'm having trouble using the latex thingy ><

Let's go back to:

\(\displaystyle \frac{2}{3}=e^{1.1x}\)

Now recall that:

\(\displaystyle a=b^c\implies c=\log_b(a)\)

And so we may write:

\(\displaystyle 1.1x=\ln\left(\frac{2}{3}\right)\)

And then on dividing though by 1.1, we get:

\(\displaystyle x=\frac{\ln\left(\dfrac{2}{3}\right)}{1.1}\)

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[Velo]

Ohhh, I get it now.. I read the book wrong too :') The solution had \(\displaystyle \log_e(\frac{2}{3}) \) and not \(\displaystyle \log(\frac{2}{3})\).. I spent so much time wondering where that log had come from :') I tried redoing that exercise and the next in my notebook and I'm doing alright now x3 Thank you very much! :D