Question

Solve the equation for x, accurate to three decimal places: e^{4x} − 6e^{2x} = 16.x = 1.040x = 1.039x = 1.386x = 0.347

Answer 1

Hello!

We have the following equation:

$e^{4x}−6e^{2x}=16$

First, we can rewrite this expression to obtain the same base in both variables, look:

$(e^{2x})^2−6(e^{2x})=16$$(e^{2x})^2−6(e^{2x})=16$$(e^{2x})^2−6(e^{2x})=16$

To make it easier, we can replace where's e^{2x} with any letter. I'll replace it with s, look:

$\begin{gathered} (e^{2x})^2-6(e^{2x})=16 \\ \\ \left(e^{2x}\right)=s \\ \\ s^2−6s=16 \end{gathered}$

Now, let's solve this quadratic equation:

$\begin{gathered} s^2-6s=16 \\ s^2-6s-16=0 \end{gathered}$$\begin{gathered} s=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(-16)}}{2\times1} \\ \\ =\frac{6\pm\sqrt{36^+64}}{2\times1}=\frac{6\pm\sqrt{100}}{2}=\frac{6\pm10}{2} \\ \\ s^{\prime}=\frac{6+10}{2}=\frac{16}{2}=8 \\ \\ s^{\prime}^{\prime}^=\frac{6-10}{2}=-\frac{4}{2}=-2 \end{gathered}$

At this moment, we obtained two possible values for this equation: -2 and 8.

Now, let's cancel the replacement that we did before. s will turn back to be e^{2x}.

$\begin{gathered} e^{2x}=-2\notin\mathrm{R} \\ e^{2x}=8\text{ }\in R \\ \end{gathered}$

So, let's solve the second line:

$\begin{gathered} e^{2x}=8 \\ \ln(e^{2x})=\ln(8) \\ 2x=\ln(2^3) \\ 2x=3\ln(2) \\ x=\frac{3\ln(2)}{2} \end{gathered}$

Using a calculator, we will obtain x = 1.03972.

As we have to round to three decimal places, the answer will be x = 1.040.