Question

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

Answer 1

Hint: In these type of questions, we will use the properties of logarithms, as logarithms and exponents are related to each other, i.e., logarithms are inverse of the exponents, so we will use the properties of the exponents, so now by definition the logarithm of a given number $x$ is the exponent to another number, with the base b, must be raided to produce.

Complete step-by-step answer:
Given equation is ${\log _{x + 1}}{\left( {{x^2} + x - 6} \right)^2} = 4$,
Now we have to solve the given equation, and get the sum of all these solutions after multiplying it with 2.
Now we have to change the equation using logarithmic property as follows,
$ \Rightarrow {\log _{x + 1}}\left( {{x^2} + x - 6} \right) = 4$,
Now using the property, \[{\log _a}{b^x} = x \cdot {\log _a}b\],
\[ \Rightarrow 2{\log _{x + 1}}\left( {{x^2} + x - 6} \right) = 4\],
Again simplifying we get,
\[ \Rightarrow {\log _{x + 1}}\left( {{x^2} + x - 6} \right) = 2\],
Now we can write as it fulfils each condition we get,
\[ \Rightarrow \left( {x + 1} \right) \ne 1\],
By simplifying we get,
\[ \Rightarrow x \ne 0\],
The property used in the above is the condition for a logarithmic function to exist is that input must be greater than 0,
\[ \Rightarrow x + 1 > 0\],
\[ \Rightarrow x > - 1\],
$ \Rightarrow x \in \left( { - 1,\infty } \right)$,
Now the condition is that the base of the logarithmic function should not be equal to 1.
\[ \Rightarrow {x^2} + x - 6 > 0\],
\[ \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) > 0\],
\[ \Rightarrow x \in \left( { - \infty , - 3} \right) \cup \left( {2,\infty } \right)\],
Now the logarithmic property that we use is that the base must be greater than 0, so now using the intersection of all the three sets we get, \[x \in \left( {2,\infty } \right)\],
Now solving the equation we get,
\[ \Rightarrow {\log _{x + 1}}\left( {{x^2} + x - 6} \right) = 2\],
Now taking the base to the R.H.S, we get,
\[ \Rightarrow \left( {{x^2} + x - 6} \right) = {\left( {x + 1} \right)^2}\],
By simplifying we get,
\[ \Rightarrow \left( {{x^2} + x - 6} \right) = {x^2} + 2x + 1\],
By eliminating like terms we get,
\[ \Rightarrow x - 6 = 2x + 1\],
Now taking x terms to one side, we get,
\[ \Rightarrow x = - 7\],
As the value of\[x\] does not lie in the domain of\[x\], which is the set A, hence there will be no solution which means the sum of its solution is 0 and its twice will also be 0.

The given equation has no solutions.

Note:
In these types of questions students should know the properties of the logarithms and should also know where to apply these properties to get the required solution, and also should know how to find the solution set multiple sets which is done by taking the intersection of the sets.