Question

The given equation is quadratic in form. Solve the following equation and give exact solutions.2 e^2x+e^x=15The equation rewritten as a quadratic in standard form using u-substitutions is?

Answer 1

In order to rewrite this equation into a quadratic equation, let's substitute the term e^x by a new variable: y. So we have:

`\begin{gathered} 2e^(2x)+e^x=15 \\ 2(e^x)^2+e^x=15 \\ 2y^2+y=15 \\ 2y^2+y-15=0 \end{gathered}`

Now, solving this equation using the quadratic formula, we have:

`\begin{gathered} a=2,b=1,c=-15 \\ y_1=\frac{-b+\sqrt[]{b^2-4ac}}{2a}=\frac{-1+\sqrt[]{1+120}}{4}=(-1+11)/(4)=(5)/(2) \\ y_2=\frac{-b-\sqrt[]{b^2-4ac}}{2a}-(-1-11)/(4)=-3 \end{gathered}`

Now, going back to the substitution y = e^x, we have:

`\begin{gathered} e^x=(5)/(2) \\ \ln (e^x)=\ln ((5)/(2)) \\ x=0.9163 \\ \\ e^x=-3 \\ x\text{ does not exist} \end{gathered}`

So the solution to the initial equation is x = ln(5/2)