Question

If \[{{\log }_{(x+1)}}{{({{x}^{2}}+x-6)}^{2}}=4\] , then the value(s) of twice the sum of all possible values of x is/are

Answer 1

HINT: - Before solving this question, we must know the properties of a logarithmic function which are as follows
The necessary condition for a logarithmic function to exist is that the base as well as the input must be greater than 0.
Another important condition that needs to be fulfilled is that the base of the logarithmic function must not be equal to 1.
In such questions, we try to fulfill all the necessary conditions and then we try to find the solutions to the equation.

Complete step-by-step answer:
As mentioned in the question, we have to find the solutions of the given equation and then get the sum of all these solutions after multiplying it with 2.
Now, first of all, we can change the equation using logarithmic property as follows
 \[\begin{align}
  & \Rightarrow {{\log }_{(x+1)}}{{({{x}^{2}}+x-6)}^{2}}=4 \\
 & \left[ {{\log }_{(a)}}{{(b)}^{x}}=x\cdot {{\log }_{(a)}}(b) \right] \\
 & \Rightarrow 2{{\log }_{(x+1)}}({{x}^{2}}+x-6)=4 \\
 & \Rightarrow {{\log }_{(x+1)}}({{x}^{2}}+x-6)=2 \\
\end{align}\]
Now, on fulfilling each and every condition, we can write as follows
\[\begin{align}
  & \Rightarrow (x+1)\ne 1 \\
 & \Rightarrow x\ne 0 \\
\end{align}\]
The property used above is the condition that needs to be fulfilled is that the base of the logarithmic function must not be equal to 1.

\[\begin{align}
  & \Rightarrow ({{x}^{2}}+x-6)>0 \\
 & \Rightarrow (x+3)(x-2)>0 \\
 & \Rightarrow x\in (-\infty ,-3)\cup (2,\infty ) \\
\end{align}\]
The property that is used above is the condition for a logarithmic function to exist is that the input must be greater than 0.
\[\begin{align}
  & \Rightarrow x+1>0 \\
 & \Rightarrow x>-1 \\
 & \Rightarrow x\in (-1,\infty ) \\
\end{align}\]
The property that is used above is that the condition for a logarithmic function to exist is that the base must be greater than 0.
Now, on taking the intersection of all the sets that we are getting as mentioned above, we get the following set
\[A=x\in \left( 2,\infty \right)\]
Now, on solving the expression, we get the following result
\[\begin{align}
  & \Rightarrow {{\log }_{(x+1)}}({{x}^{2}}+x-6)=2 \\
 & \Rightarrow ({{x}^{2}}+x-6)={{(x+1)}^{2}}={{x}^{2}}+1+2x \\
 & \Rightarrow x=-7 \\
\end{align}\]
As this value of x does not lie in the domain of x which is the set A, hence, the equation has no solution which means the sum of its solutions is 0 and the twice of it is also 0.

NOTE: - The students can make an error if they don’t know the properties of a logarithmic function and also the necessary conditions for it as without knowing them one could never get to the right solution.
It is also important to know how to find the solution set of multiple sets which is done by taking the intersection of those sets as mentioned below
Let A, B and C be the sets of which the solution set is to be obtained. Hence,
\[Solution\ set=A\bigcap B\bigcap C\]
Also, in such questions, calculation mistakes are very common, so, to get to the right answer, one should try to be extra careful while solving these questions.