Question

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

Explanation:

Step1: Isolate the exponential term

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

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:
\(\ln(e^{2x})=\ln(3)\)
Using the logarithm property \(\ln(e^a)=a\), we get:
\(2x=\ln(3)\)

Step4: Solve for \(x\)

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

Answer:

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