\(|x-3|^{3x^2 – 10x + 3} = 1\). Find \(x\).

There can be three cases:

CASE I : \(3x^2 – 10x + 3 = 0\) and \(|x-3| \neq 0\)

\(3x^2 – 10x + 3 = 0\) and \(x-3 \neq 0\)
\(\Rightarrow 3x^2 – 9x – x + 3 = 0\) and \(x \neq 3\)
\(\Rightarrow 3x(x – 3) – 1(x – 3) = 0\) and \(x \neq 3\)
\(\Rightarrow (3x – 1)(x – 3) = 0\) and \(x \neq 3\)
\(x = \frac{1}{3}\) or \(x = 3\) and \(x \neq 3\)

Therefore from CASE I , \(x = \frac{1}{3}\)

CASE II : \(|x – 3| = 1\)

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\(x\;-\;3 \pm 1\)
\(\Rightarrow x = 4\) OR \(x = 2\)

From CASE II , \(x = 4 , x = 2\)

CASE III : \(|x – 3| = -1\) and \(3x^2 – 10x + 3\) is an even number or is equal to 0

But \(|x – 3| = -1\) is not possible.

Hence we get \(x \in \{\frac{1}{3} , 4 , 2 \}\)