What did I do wrong with this logarithmic equation?

Question

$$e^{3x-2}e^{-x}=4e, \ \text{round to the nearest thousandths}$$

I keep getting $x\approx2.884$ but the answer is $x\approx2.193$. What am I doing wrong? Here is my work:

\begin{align*} e^{3x-2}e^{-x}&=4e \\ e^{2x-2}&=4e \\ 2x-2\ln(e)&=\ln(4e) \\ 2x-2&=\ln(4e)\\ 2x&=\ln(4e)+2 \\ x&=\frac{\ln(4e)+2}{2} \\ x&\approx2.884 \end{align*}

I've tried looking at $e^{-x}$ as $\frac{1}{e^x}$ or calculating the exact value of some of the simpler natural logs, but I keep getting the same answer. I feel like the mistake I'm making is so ridiculously obvious but I'm just not seeing it.

UPDATE: I discovered my mistake was an error of notation. I did not properly include parentheses on the last step on my calculator, and therefore my calculator assumed I was computing $\frac{\ln(4)e+2}{2}$

Answer 1

The calculation (last step) step is wrong.

$$ x = \frac{\ln(4e) + 2}{2} = \frac{\ln 2^{2} + \ln e + 2}{2} = \frac{2\ln 2 + 3}{2} = \ln 2 + \frac{3}{2} \approx 2.193 $$

(Also, you should have used parentheses the third step down.)

Answer 2

As indicated by the third line of your computation, you were careless with your parentheses. What you computed is $$ \frac{\ln(4)\cdot\mathrm{e} + 2}{2} \approx 2.884. $$ My guess is that what you typed into the calculator was something like

 ( ln 4 * e + 2 ) / 2

or possibly

 ln 4 * e + 2 / 2

depending on what calculator or computer program you are using.

Your first edit also does not fix the issue. Your calculator was not assuming that you meant $$ \frac{\ln(4\mathrm{e} + 2)}{2} \approx 2.996. $$ What you meant to calculate was $$ \frac{\ln(4\mathrm{e}) + 2}{2} \approx 2.193. $$ On most calculators, this would be something like

 ( ln ( 4 * e ) + 2 ) / 2

Parentheses matter. Be careful with them.