Question

How do you solve $2\ln (x + 3) = 0$ and find any extraneous solutions?

Answer 1

Hint: In this question we will use the properties of logarithm and apply it to both the sides of the given equation to simplify and solve it for the value of $x$. Then we put the value in the given equation, we will conclude the required answer.

Complete step-by-step solution:
We have the question given us to as:
$ \Rightarrow 2\ln (x + 3) = 0$
Now we know the general property of logarithm that $a\log b = \log {b^a}$ therefore on applying this property of log on the given equation we get:
$ \Rightarrow \ln {(x + 3)^2} = 0$
Now since in the left-hand side of the equation we have the term expression in the form of logarithm and on the right-hand side we don’t, we know that $\ln 1 = 0$, therefore on substituting the value of $0$ as $\ln 1$ in the right-hand side we get:
$ \Rightarrow \ln {(x + 3)^2} = \ln 1$
Now since the base to the exponent is same on both the sides, we can just remove and write the terms as:
\[ \Rightarrow {(x + 3)^2} = 1\]
Now on taking the square root on both the sides we get:
$ \Rightarrow x + 3 = \pm 1$
Therefore, we have $2$equations and they can be written as:
$ \Rightarrow x + 3 = 1$ and
$ \Rightarrow x + 3 = - 1$
Therefore, we have the solutions as:
$ \Rightarrow x = 1 - 3$ and
$ \Rightarrow x = - 1 - 3$
On simplifying we get:
$ \Rightarrow x = - 2$ and
 $ \Rightarrow x = - 4$
Which are the solutions of the term
Now to check whether the solutions extraneous we will re-substitute the solutions in the equation.
On substituting $x = - 2$ in the term $2\ln (x + 3) = 0$ we get:
$ \Rightarrow 2\ln ( - 2 + 3) = 0$
This can be simplified as:
$\ln 1 = 0$, which is true therefore $x = - 2$ it is not an extraneous solution.
Now on substituting $x = - 4$ in the term $2\ln (x + 3) = 0$ we get:
$ \Rightarrow 2\ln ( - 4 + 3) = 0$
This can be simplified as:
$ \Rightarrow 2\ln ( - 1) = 0$

Since the log of negative numbers do not exist, the given statement is not true therefore $x = - 4$ it is an extraneous solution.

Note: It is to be noted that the square root of a positive number has $2$ roots which are positive and negative respectively.
Logarithm are the form of ${\log _a}b$ where $a$ is the base of the log, commonly used bases are $10$ and $e$,
${\log _e}a$ is generally written as: $\ln a$.