To solve the equation \( 10 e^{2x} - 5 = 23 e^{x} \) for \( x \), first, let's rearrange the equation:
- Move all terms to one side:
\[ 10 e^{2x} - 23 e^{x} - 5 = 0 \]
- To simplify this expression, we can make a substitution. Let \( y = e^{x} \). Then \( e^{2x} = y^2 \), and our equation becomes:
\[ 10 y^2 - 23 y - 5 = 0 \]
- 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 \( c = -5 \).
- Now calculate the discriminant:
\[ b^2 - 4ac = (-23)^2 - 4 \cdot 10 \cdot (-5) = 529 + 200 = 729 \]
- Now apply the quadratic formula:
\[ y = \frac{23 \pm \sqrt{729}}{20} = \frac{23 \pm 27}{20} \]
Calculating the two possible values of \( y \):
- First value of \( y \):
\[ y = \frac{23 + 27}{20} = \frac{50}{20} = 2.5 \]
- Second value of \( y \):
\[ y = \frac{23 - 27}{20} = \frac{-4}{20} = -0.2 \]
Since \( y = e^x \) must be positive, we discard \( y = -0.2 \).
- Now, set \( y = 2.5 \):
\[ e^x = 2.5 \]
- 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:
\[ \ln \left(\frac{5}{2}\right) \]