0
$\begingroup$

Solve, correct to 3 significant figures, the equation

$$e^x + e^{2x} = e^{3x}$$

How do you go about doing these type of questions?

$\endgroup$
  • 1
    $\begingroup$ Set $t=e^x$ and see what type of equation you get. Hint: $\log\phi$. $\endgroup$ – Yves Daoust Apr 2 '17 at 10:54
1
$\begingroup$

Hint : Since $e^x \neq 0$, Divide the equation by $e^x$ and let $e^x=t$. This will give you a quadratic equation in $t$. The solve for $t$, and there you will get $x = \ln t$

When you solve for $t$, reject negative value (since $e^x > 0 ~\forall x \in \mathbb{R}$) and you will get $x \approx 0.481$

$\endgroup$
  • $\begingroup$ $e^{2x}= (e^x)^2=(t)^2$ $\endgroup$ – Jaideep Khare Apr 2 '17 at 10:58
1
$\begingroup$

Setting $$t=e^x$$ we get $$t+t^2=t^3$$ or $$t(t^2-t-1)=0$$ can you solve this?

$\endgroup$
0
$\begingroup$

Define $$t:=e^x.$$

Substituting into your equation, $$t+t^2=t^3.$$

So one obvious solution is $t_1=0.$. If we assume $t\neq 0$, then we can divide everything by $t$ to get $$1+t=t^2.$$ Using the quadratic formula, we also get the following two solutions: $$t_{2,3}=\frac{1\pm\sqrt{1-4(1)(-1)}}{2}.$$ So our solutions are $$x_1=\ln(0);\qquad x_2=\ln(\frac{1+\sqrt{5}}{2});\qquad x_3=\ln(\frac{1-\sqrt{5}}{2}).$$

Note that $\ln(0)$ is undefined and that the argument of the logarithm in the third one is negative, so we only accept the second solution.

$\endgroup$
  • 1
    $\begingroup$ Also, not third one it is $\ln (<0)$ $\endgroup$ – Jaideep Khare Apr 2 '17 at 11:04
  • 1
    $\begingroup$ Perfect, got it! $\endgroup$ – Riduan Gonzalez Apr 2 '17 at 11:04
  • $\begingroup$ @JaideepKhare thanks - fixed! $\endgroup$ – man_in_green_shirt Apr 2 '17 at 11:35

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.