2
$\begingroup$

What are the two solutions to this equation? $$e^{2x}-21e^x+110=0$$ I'm stumped! Taking $\log$ of both sides doesn't seem to work?

$\endgroup$
  • 5
    $\begingroup$ Set $y=e^x$ and solve as a regular quadratic (since $e^{2x}=y^2$). Once you have $y$, take $x=ln(y)$. $\endgroup$ – user2825632 Sep 17 '16 at 4:27
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Sep 17 '16 at 4:30
  • $\begingroup$ How about a u-substitution with $u=e^{x} $. $\endgroup$ – Joe Sep 17 '16 at 4:54
5
$\begingroup$

Hint: Write the equation as $$(e^x)^2-21(e^x)+110=0$$ and solve for $e^x$. Then solve the resulting equations for $x$.

You can simply factor as $$(e^x-10)(e^x-11)=0$$ and go from there.

So now you should be able to solve $ae^{2x}+be^x+c=0$.

$\endgroup$
0
$\begingroup$

Set

$y=e^{x}$, so $y^{2}=e^{2x}$, and we had $e^{2x}-21e^{x}+110=0$

So: $$y^{2}-21y+110=0$$

This is Quadratic Equation and if we solve this (for example with Quadratic formula), two values will be calculated for $y$:

$$y_{1}=10\\ y_{2}=11$$

At first we set $y=e^{x}$, so $x=ln(y),\ y>0$. $y_1$ and $y_2$ are positive ($y_1, y_1>0$) so there are two values for $x $, $x_1 $ and $ x_2$:

$$x_1 = \ln(y_1) = \ln(10)$$ $$x_2 = \ln(y_2) = \ln(11)$$

$\endgroup$

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.