Multiple choice

Number of real roots of the equation ${ 3 }^{ {\log }_{ 3 }\left( { x }^{ 2 }-4x+3 \right) }=(x-3)$ is:

  1. $0$
  2. $1$
  3. $2$
  4. $3$
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A Correct answer
Explanation

The equation 3^(log3(x^2 - 4x + 3)) = x - 3 simplifies to x^2 - 4x + 3 = x - 3, which is x^2 - 5x + 6 = 0. Factoring gives (x - 2)(x - 3) = 0, so x = 2 or x = 3. However, the logarithm argument x^2 - 4x + 3 must be positive; for x = 2, 4 - 8 + 3 = -1 (invalid), and for x = 3, 9 - 12 + 3 = 0 (invalid). Thus, there are no real roots.

AI explanation

Using the identity a^(log_a(b)) = b, the equation becomes x^2 - 4x + 3 = x - 3, provided that x^2 - 4x + 3 > 0 and x > 3. This quadratic simplifies to x^2 - 5x + 6 = 0, yielding potential roots x = 2 and x = 3. Neither root is strictly greater than 3 as required by the domain restrictions, so both must be discarded. The result is 0.

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