Question

what is the solution to $3 + 4e^{x + 1} = 11$?
\\(\bigcirc\\) $x = \ln 2 - 1$
\\(\bigcirc\\) $x = \ln 2 + 1$
\\(\bigcirc\\) $x = \frac{1}{e}$
\\(\bigcirc\\) $x = \frac{e + 2}{e}$

Explanation:

Step1: Isolate the exponential term

Subtract 3 from both sides of the equation \(3 + 4e^{x + 1}=11\).
\(4e^{x + 1}=11 - 3 = 8\)

Step2: Solve for the exponential part

Divide both sides by 4:
\(e^{x + 1}=\frac{8}{4}=2\)

Step3: Take the natural logarithm

Take the natural log of both sides: \(\ln(e^{x + 1})=\ln(2)\)
Since \(\ln(e^{a}) = a\), we get \(x + 1=\ln(2)\)

Step4: Solve for x

Subtract 1 from both sides: \(x=\ln(2)-1\)

Answer:

\(x = \ln2 - 1\) (the first option: \(x=\ln2 - 1\))