6
$\begingroup$

I recently took my Honors Pre-Calc. exam. Wanting to get as many points as I possibly could, I circled and skipped several problems. However, I forgot to do this one, and wanted to know if I'm on the right track, because the answer I get for x doesn't work correctly. I've come up with a separate solution that comes close, but is off by .4 as well. \begin{align*} e^{2x} - 7e^x + 12 &= 0\\
e^{2x} - 7e^x &= -12 \\ e^x(1^2 - 7) &= -12\\
e^x &= \frac{-12}{1^2 - 7}\\
e^x &= 2\\
x &= \ln 2 \end{align*}

Putting this back into the equation doesn't work. I think I might have a problem with how I'm factoring, but I can't seem to fix it. The only other way I've tried comes closer, but still gives me a wrong answer.

Your help and suggestions are appreciated :).

$\endgroup$
  • 3
    $\begingroup$ $e^{2x} \neq e^x$ $\endgroup$ – Raphael Dec 18 '10 at 16:57
  • $\begingroup$ $ e^{2x}=e^x\cdot e^x\neq e^x\text{ for }x\neq0$ $\endgroup$ – Hakim Mar 27 '14 at 19:29
19
$\begingroup$

HINT $\rm\quad e^{\:2\:x} = (e^x)^2\:.\ $ Thus for $\rm\ y = e^x\ $ the equation is $\rm\ y^2 - 7\ y + 12 = 0\:$

$\endgroup$
  • $\begingroup$ Alright, that helps. Thank you. $\endgroup$ – Insomaniacal Dec 18 '10 at 17:55
4
$\begingroup$

Your error is in the third line. $e^{2x}$ is not equal to $e^x$ times $1^2$ (that would make $e^{2x}$ equal to $e^x\times 1 = e^x$). That line would read $$e^x(e^x - 7) = -12,$$ and that won't really help you. Bill's solution is the correct approach.

$\endgroup$
1
$\begingroup$

If you're good at factoring quadratic polynomials by staring at them (and I've seen lots of people who are surprisingly adept at that even if some of them can't do much else), then you can do this: $$ e^{2x} - 7 e^x + 12 = 0 $$ $$ (e^x - 3)(e^x - 4) = 0 $$ $$ e^x=3 \text{ or }e^x=4. $$ $$ x = \log_e 3\text{ or } x = \log_3 4. $$

$\endgroup$
  • 1
    $\begingroup$ The point of my posting hint was to leave some work for the reader. Why finish it? $\endgroup$ – Bill Dubuque Dec 7 '11 at 0:33
  • 5
    $\begingroup$ If the reader were going to wade through year-old posts to solve a quadratic in $e^x$, wouldn't that qualify as "some work" in itself? $\endgroup$ – The Chaz 2.0 Dec 7 '11 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.