Question

which of the following is the solution of $5e^{2x}-4 = 11$?
$\bigcirc$ $x = \ln 3$
$\bigcirc$ $x = \ln 27$
$\bigcirc$ $x = \frac{\ln 3}{2}$
$\bigcirc$ $x = \frac{3}{\ln 3}$

Explanation:

Step1: Isolate the exponential term

Start with the equation \( 5e^{2x} - 4 = 11 \). Add 4 to both sides:
\( 5e^{2x} = 11 + 4 \)
\( 5e^{2x} = 15 \)

Step2: Solve for \( e^{2x} \)

Divide both sides by 5:
\( e^{2x} = \frac{15}{5} \)
\( e^{2x} = 3 \)

Step3: Take the natural logarithm

Take \( \ln \) of both sides (since \( \ln(e^y) = y \)):
\( \ln(e^{2x}) = \ln(3) \)
\( 2x = \ln(3) \)

Step4: Solve for \( x \)

Divide both sides by 2:
\( x = \frac{\ln(3)}{2} \)

Answer:

\( x = \frac{\ln 3}{2} \) (the third option)