The real roots of the equation $5^{\log_5 (x^2-4x+5)}=x-1$ are
- $1$ and $2$
- $2$ and $3$
- $3$ and $4$
- $4$ and $5$
Reveal answer
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B
Correct answer
Explanation
The equation 5^(log5(x^2-4x+5)) = x-1 simplifies to x^2-4x+5 = x-1, which rearranges to x^2-5x+6 = 0. Factoring gives (x-2)(x-3) = 0, so the roots are 2 and 3. Both roots satisfy the domain condition of the logarithm (x^2-4x+5 > 0).
AI explanation
Using the identity a^(log_a(x)) = x, the left side simplifies to x^2 - 4x + 5, making the equation x^2 - 4x + 5 = x - 1. Rearranging yields x^2 - 5x + 6 = 0, which factors to (x - 2)(x - 3) = 0. Checking the conditions x - 1 > 0 and x^2 - 4x + 5 > 0 for x = 2 and x = 3 shows both values are valid, giving the roots 2 and 3.