Question

Solve e superscript 2x baseline plus 3e superscript x baseline minus 28 equals 0.
(1 point)
Responses

x equals ln left parenthesis negative 7 right parenthesis or x = ln 4
Image with alt text: x equals ln left parenthesis negative 7 right parenthesis or x = ln 4

x = ln 4
x = ln 4

x equals ln left parenthesis negative 4 right parenthesis or x = ln 7
Image with alt text: x equals ln left parenthesis negative 4 right parenthesis or x = ln 7

x = ln 7
x = ln 7

Answer 1

To solve the equation \( e^{2x} + 3e^x - 28 = 0 \), we can use a substitution. Let \( y = e^x \). Then \( e^{2x} = y^2 \), and our equation becomes:

\[
y^2 + 3y - 28 = 0
\]

This is a quadratic equation in standard form \( ay^2 + by + c = 0 \), where \( a = 1, b = 3, c = -28 \).

We can use the quadratic formula:

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

Plugging in the values:

\[
y = \frac{{-3 \pm \sqrt{{3^2 - 4(1)(-28)}}}}{2(1)}
\]
\[
y = \frac{{-3 \pm \sqrt{{9 + 112}}}}{2}
\]
\[
y = \frac{{-3 \pm \sqrt{121}}}{2}
\]
\[
y = \frac{{-3 \pm 11}}{2}
\]

Calculating the two possible values for \( y \):

1. \( y = \frac{{-3 + 11}}{2} = \frac{8}{2} = 4 \)
2. \( y = \frac{{-3 - 11}}{2} = \frac{-14}{2} = -7 \)

Since \( y = e^x \) and \( e^x \) can only be positive, we discard \( y = -7 \) and keep \( y = 4 \).

Now, substituting back for \( y \):

\[
e^x = 4
\]

Taking the natural logarithm of both sides:

\[
x = \ln(4)
\]

Thus, the solution to the original equation is:

\[
x = \ln(4)
\]

The other proposed solutions (ln(-7) or ln(-4) or ln(7)) do not apply, as logarithms of negative numbers are undefined.

**Final Answer:**
\( x = \ln(4) \)