Question

How do you solve $\ln \left( x-7 \right)=2$?

Answer 1

We solve the given equation using the identity formula of logarithm ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. We decide on the base of the logarithmic base. The base of $\ln$ in general cases is always $e$. The main step would be to eliminate the logarithm function and keep only the linear equation of x. We solve the linear equation with the help of basic binary operations.

Complete step-by-step solution:
We take the logarithmic identity for the given equation $\ln \left( x-7 \right)=2$ to find the solution for x.
We have $\ln a={{\log }_{e}}a$.
We have a single equation of logarithm and a constant on the opposite sides of the equation.
Therefore, $\ln \left( x-7 \right)={{\log }_{e}}\left( x-7 \right)=2$.
Now we have to eliminate the logarithm function to find the value of x.
We know ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. Applying the rule in case of ${{\log }_{e}}\left( x-7 \right)=2$, we get
$\begin{aligned} & {{\log }_{e}}\left( x-7 \right)=2 \\\ & \Rightarrow \left( x-7 \right)={{e}^{2}} \\\ \end{aligned}$
Now we have a linear equation of x. We need to solve it.
We add 7 on both sides of the equation $\left( x-7 \right)={{e}^{2}}$ and get
$\begin{aligned} & \left( x-7 \right)={{e}^{2}} \\\ & \Rightarrow x-7+7={{e}^{2}}+7 \\\ & \Rightarrow x={{e}^{2}}+7 \\\ \end{aligned}$
Therefore, the solution for the equation $\ln \left( x-7 \right)=2$ is $x={{e}^{2}}+7$.

Note: The logarithm is used to convert a large or very small number into the understandable domain. For the theorem to work the usual conditions of logarithm will have to follow. In case the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $e$ is fixed for $\ln$.
We also need to remember that for logarithm function there has to be a domain constraint.
For any ${{\log }_{e}}a$, $a>0$. This means for $\ln \left( x-7 \right)$, $\left( x-7 \right)>0$. The simplified form is $x>7$.