Question

Solve 10 e superscript 2 x baseline minus 5 equals 23 e superscript x baseline for x.
(1 point)
Responses

ln left-parenthesis Start Fraction 1 over 5 End Fraction right-parenthesis
Image with alt text: ln left-parenthesis Start Fraction 1 over 5 End Fraction right-parenthesis

ln left-parenthesis Start Fraction 5 over 2 End Fraction right-parenthesis
Image with alt text: ln left-parenthesis Start Fraction 5 over 2 End Fraction right-parenthesis

ln left-parenthesis Start Fraction 5 over 13 End Fraction right-parenthesis
Image with alt text: ln left-parenthesis Start Fraction 5 over 13 End Fraction right-parenthesis

no solution

Answer 1

To solve the equation \( 10 e^{2x} - 5 = 23 e^{x} \) for \( x \), first, let's rearrange the equation:

1. Move all terms to one side:

\[
10 e^{2x} - 23 e^{x} - 5 = 0
\]

2. To simplify this expression, we can make a substitution. Let \( y = e^{x} \). Then \( e^{2x} = y^2 \), and our equation becomes:

\[
10 y^2 - 23 y - 5 = 0
\]

3. Next, we can solve this quadratic equation using the quadratic formula:

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

where \( a = 10 \), \( b = -23 \), and \( c = -5 \).

4. Now calculate the discriminant:

\[
b^2 - 4ac = (-23)^2 - 4 \cdot 10 \cdot (-5) = 529 + 200 = 729
\]

5. Now apply the quadratic formula:

\[
y = \frac{23 \pm \sqrt{729}}{20} = \frac{23 \pm 27}{20}
\]

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

1. First value of \( y \):

\[
y = \frac{23 + 27}{20} = \frac{50}{20} = 2.5
\]

2. Second value of \( y \):

\[
y = \frac{23 - 27}{20} = \frac{-4}{20} = -0.2
\]

Since \( y = e^x \) must be positive, we discard \( y = -0.2 \).

6. Now, set \( y = 2.5 \):

\[
e^x = 2.5
\]

7. Take the natural logarithm of both sides:

\[
x = \ln(2.5)
\]

Thus, the solution is:

\[
\ln(2.5)
\]

Now looking at the provided responses, we can multiply both the numerator and denominator by 2 for clarity:

\[
\ln \left(\frac{5}{2}\right)
\]

This means that the correct answer is:

\[
\ln \left(\frac{5}{2}\right)
\]