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

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 $\frac{\ln(4)e+2}{2}$

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    $\begingroup$ In my calculator, $(\ln(4e) + 2) / 2$ is approximately $2.193$. I don't see any mistake in your calculation. $\endgroup$
    – paulinho
    Jul 13, 2019 at 2:43
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    $\begingroup$ In your third line of computation, you presumably mean $(2x-2)\ln(\mathrm{e})$? $\endgroup$
    – Xander Henderson
    Jul 13, 2019 at 2:43
  • $\begingroup$ @XanderHenderson He/she could have meant that, but it's possible he could have meant what he/she wrote. Nonetheless, it doesn't matter. $\endgroup$
    – paulinho
    Jul 13, 2019 at 2:45
  • $\begingroup$ I probably should utilize parentheses more often, but yes that is what I meant. It doesn't make much of a difference though, considering $\ln (e)=1$. $\endgroup$
    – Lex_i
    Jul 13, 2019 at 2:47
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    $\begingroup$ How do you get $\frac {\ln 4e + 2}2 = 2.884$? $e = 2.7183$ and $4e = 10.8732$ and $\ln 4e = 2.3863$ and $\frac {\ln 4e + 2}2 = 2.1932$. But you should have done $\ln 4e = \ln 4 + \ln e = \ln 4 + 1$. And $\ln 4 = 1.386$. $\endgroup$
    – fleablood
    Jul 13, 2019 at 2:53

2 Answers 2

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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.)

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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.

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