what is the solution to $3 + 4e^{x + 1}=11$?\n$x=ln2 - 1$\n$x=ln2 + 1$\n$x=\frac{1}{e}$\n$x=\frac{e + 2}{e}$

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: Use natural - logarithm property

Take the natural logarithm of both sides. Since $\ln(e^{a})=a$, we have $\ln(e^{x + 1})=\ln(2)$. $x + 1=\ln(2)$.

Step4: Solve for x

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

Answer:

$x=\ln2 - 1$