Question

what is the value of x in the equation below?
1 + 2e^{x + 1} = 9
\bigcirc\\ x = \log 4 - 1
\bigcirc\\ x = \log 4
\bigcirc\\ x = \ln 4 - 1
\bigcirc\\ x = \ln 4

Explanation:

Step1: Isolate the exponential term

Subtract 1 from both sides of the equation \(1 + 2e^{x + 1}=9\).
\(2e^{x + 1}=9 - 1 = 8\)

Step2: Solve for the exponential part

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

Step3: Take the natural logarithm of both sides

Since the base of the exponential is \(e\), we use the natural logarithm (\(\ln\)). Taking \(\ln\) of both sides:
\(\ln(e^{x + 1})=\ln(4)\)
Using the property \(\ln(e^{a}) = a\), we get:
\(x + 1=\ln(4)\)

Step4: Solve for \(x\)

Subtract 1 from both sides:
\(x=\ln(4)-1\)

Answer:

\(x = \ln4 - 1\) (the third option: \(x=\ln4 - 1\))