To solve for [tex]\( x \)[/tex] in the equation [tex]\( 8 \cdot e^{x-1} = 56 \)[/tex], we can follow these steps:
1. Divide both sides by 8 to simplify the equation:
[tex]\[
e^{x-1} = \frac{56}{8}
\][/tex]
[tex]\[
e^{x-1} = 7
\][/tex]
2. Take the natural logarithm (ln) on both sides to eliminate the exponent:
[tex]\[
\ln(e^{x-1}) = \ln(7)
\][/tex]
3. Use the property of logarithms, [tex]\(\ln(e^a) = a\)[/tex], to simplify the left side:
[tex]\[
x-1 = \ln(7)
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \ln(7) + 1
\][/tex]
So, the correct steps would be:
[tex]\[
\begin{aligned}
8 \cdot e^{x-1} & =56 \\
e^{x-1} & =7 \\
\ln(e^{x-1}) & = \ln(7) \\
(x-1) & = \ln(7) \\
x & = \ln(7) + 1
\end{aligned}
\][/tex]
Therefore, the correct set of steps for solving the equation is:
[tex]\[
\begin{aligned}
e^{x-1} & =7 \\
\ln(e^{x-1}) & = \ln(7) \\
(x-1) \ln(e) & = \ln(7) \\
x-1 & = \ln(7) \\
x & = \ln(7) + 1
\end{aligned}
\][/tex]
The answer that matches these steps in your options is:
[tex]\[
\begin{aligned}
8 \cdot e^{x-1} & =56 \\
e^{x-1} & =7 \\
\ln(e^{x-1}) & = \ln(7) \\
(x-1) & = \ln(7) \\
x & = \ln(7) + 1
\end{aligned}
\][/tex]
