what is the value of x in the equation below?\n1 + 2e^{x + 1}=9\nx = log4 - 1\nx = log4\nx = ln4 - 1\nx = ln4

Explanation:

Step1: Isolate the exponential term

Subtract 1 from both sides of the equation $1 + 2e^{x + 1}=9$. $2e^{x + 1}=9 - 1=8$. Then divide both sides by 2: $e^{x + 1}=\frac{8}{2}=4$.

Step2: Apply natural - logarithm

Take the natural logarithm of both sides of the equation $e^{x + 1}=4$. Since $\ln(e^{a})=a$, we have $\ln(e^{x + 1})=\ln4$. So $x + 1=\ln4$.

Step3: Solve for x

Subtract 1 from both sides to get $x=\ln4-1$.

Answer:

$x=\ln4 - 1$