# Solve exponential equation different powers

Solve, correct to 3 significant figures, the equation

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

How do you go about doing these type of questions?

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

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 $x \approx 0.481$

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

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

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.

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