https://file.aiquickdraw.com/mathbot/file/1728640516369mbnnpp1k.jpg

The equation shown in the image is:

\[
e^{2x} - e^x - 20 = 0
\]

Let's solve this step-by-step:

### Step 1: Substitution

We can simplify this by using a substitution. Let \( y = e^x \). This gives us the following equation:

\[
y^2 - y - 20 = 0
\]

### Step 2: Solve the quadratic equation

This is now a quadratic equation in \( y \). We can solve it using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( y^2 - y - 20 = 0 \), we have \( a = 1 \), \( b = -1 \), and \( c = -20 \). Plugging these values into the quadratic formula:

\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)}
\]
\[
y = \frac{1 \pm \sqrt{1 + 80}}{2}
\]
\[
y = \frac{1 \pm \sqrt{81}}{2}
\]
\[
y = \frac{1 \pm 9}{2}
\]

This gives two solutions for \( y \):

\[
y_1 = \frac{1 + 9}{2} = 5, \quad y_2 = \frac{1 - 9}{2} = -4
\]

### Step 3: Solve for \( x \)

Since \( y = e^x \), we now substitute back:

1. For \( y_1 = 5 \):
   \[
   e^x = 5 \implies x = \ln(5)
   \]
   Using a calculator:
   \[
   x_1 \approx 1.6094
   \]

2. For \( y_2 = -4 \), there is no real solution because \( e^x \) is always positive, so we discard this solution.

### Final Answer:
The only solution is \( x \approx 1.6094 \).

Would you like further details or have any questions? 

Here are 5 related questions:
1. How would you solve if the equation had additional terms, like \( e^{3x} \)?
2. Can quadratic substitution be applied in all exponential equations?
3. How do you handle equations involving natural logarithms and exponentials simultaneously?
4. What are the properties of the exponential function \( e^x \) that help in solving such equations?
5. How can you verify if the solution to an exponential equation is correct?

**Tip:** Always check the domain of the solution when dealing with logarithmic or exponential functions.

Solve the equation: e^{2x} - e^x - 20 = 0. (Round your answers to four decimal places.)

Learn how to solve the exponential equation e^{2x} - e^x - 20 = 0 using quadratic substitution and the quadratic formula. This problem covers important concepts such as algebraic exponential equations and logarithms, suitable for students in Grades 10-12.

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