Question

Need help knowing if I was right with .88 and 1

Answer 1

Given

$8e^{4x}-15e^{2x}+7=0$

Then solve for x:

Apply the law of exponents

$8\left(e^x\right)^4-15\left(e^x\right)^2+7=0$

Rewrite the equation with e^x = u:

$8\left(u\right)^4-15\left(u\right)^2+7=0$

And solve for u:

$u=1,\:u=\frac{\sqrt{14}}{4}$

Then, substitute e^x = u:

$\begin{gathered} e^x=1 \\ ln(e^x)=ln(1) \\ x=0 \end{gathered}$

and

$\begin{gathered} e^x=\frac{\sqrt{14}}{4} \\ ln(e^x)=ln(\frac{\sqrt{14}}{4}) \\ x=ln(\sqrt{14})-ln(4) \\ x=\frac{1}{2}ln(14)-2ln(2) \\ x=-0.067 \end{gathered}$

Answer: x = 0 and x = - 0.067