Determine whether or not $x = \frac { 1 } { 3 } \left( e ^ { - 3 } + 1 \right) \text { is a solution to } \ln ( 3 x - 1 ) = - 3 \text {. }$

Question

Quizplus · Accepted Answer

To verify if $x = \frac { 1 } { 3 } \left( e ^ { - 3 } + 1 \right)$ is a solution, substitute $x$ into the equation $\ln ( 3 x - 1 ) = - 3$. Simplifying gives $\ln ( e ^ { - 3 } ) = - 3$, which is true since the natural logarithm of $e^{-3}$ is indeed $-3$.

Determine Whether or Not x=13(e−3+1) is a solution to ln⁡(3x−1)=−3. x = \frac { 1 } { 3 } \left( e ^ { - 3 } + 1 \right) \text { is a solution to } \ln ( 3 x - 1 ) = - 3 \text {. }x=31​(e−3+1) is a solution to ln(3x−1)=−3.

Determine Whether or Not $x = \frac { 1 } { 3 } \left( e ^ { - 3 } + 1 \right) \text { is a solution to } \ln ( 3 x - 1 ) = - 3 \text {. }$