Question

The solution set of the equation ${x^{{{\log }_x}{{(1 - x)}^2}}} = 9$ is?
A) {-2, 4}
B) {4}
C) {0, -2, 4}
D) None of these

Answer 1

Hint: In this question, we will use the logarithm formula ${b^{{{\log }_b}a}} = a$. Now we can easily simplify the given question and obtain a quadratic equation. At the last solve the quadratic equation to get the result.

Complete step-by-step answer:
We have;
${x^{{{\log }_x}{{(1 - x)}^2}}} = 9$
We know that: ${b^{{{\log }_b}a}} = a$
We will put the value in the above formula; we get,
${(1 - x)^2} = 9$
Simplifying the above equation, we get;
$1 + {x^2} - 2x = 9$
On converting the equation into standard quadratic equation form
$ \Rightarrow {x^2} - 2x - 8 = 0$
On splitting middle term
$ \Rightarrow {x^2} - 4x + 2x - 8 = 0$
$ \Rightarrow x(x - 4) + 2(x - 4) = 0$
Taking (x – 4) common
$ \Rightarrow (x - 4)(x + 2) = 0$
$\because $ x = − 2 and x = 4

Therefore,$x \in \{ - 2,4\} $

Note: Logarithms are the opposite phenomenon of exponential like subtraction is the inverse of addition process, and division is the opposite phenomena of multiplication. Logs “undo” exponentials. Exponent of Log Rule (A base to a Logarithm Power Rule) is ${b^{{{\log }_b}a}} = a$. The logarithm of the product is the sum of the logarithms of the factors: ${\log _b}(M.N) = {\log _b}M + {\log _b}N$. Following is some of the log properties which you must remember while solving the logarithms examples.
Logarithm of negative numbers and zero is not defined.
Let us have ${\log _a}$x = b $ \Rightarrow $ x = ${a^b}$. Now ab is always a positive number whatever be the values of a and b. So, x > 0 always. Hence ‘x’ cannot be negative or zero.
Base of a logarithm cannot be 1.
Take an example like log1 3 = b ⇒ 3 = ${1^b}$. Now this can never be true. So, the base cannot be equal to 1.
The base of a logarithm is always positive.
If no base is written then it is to be taken as 10.