If $\log_{10} (x^2 - 6x + 10) = 0$, then the value of $x$ is

### Concept & Logic This problem combines basic logarithmic definitions with solving a quadratic equation. The fundamental definition of a logarithm converts it into an exponential equation. Logarithm to Exponential Conversion: If $ \log_b(A) = C $, then $ b^C = A $ ### Step-by-Step Solution **Given:** $\log_{10}(x^2 - 6x + 10) = 0$ Convert this logarithmic equation into its equivalent exponential form. The base is $10$, the exponent is $0$, and the result is the quadratic expression: $10^0 = x^2 - 6x + 10$ Since any non-zero number raised to the power of $0$ is $1$: $1 = x^2 - 6x + 10$ Rearrange the equation to set it to zero (standard quadratic form): $x^2 - 6x + 10 - 1 = 0$ $x^2 - 6x + 9 = 0$ Factor the quadratic equation. We need two numbers that multiply to $9$ and add to $-6$. Those numbers are $-3$ and $-3$. $(x - 3)(x - 3) = 0$ $(x - 3)^2 = 0$ Take the square root of both sides: $x - 3 = 0$ $x = 3$ ### Exam Strategy & Shortcut Whenever you see a log equation equal to zero, you instantly know the argument (what's inside the parentheses) must equal $1$. So, $x^2 - 6x + 10 = 1 \Rightarrow x^2 - 6x + 9 = 0$. Recognizing $x^2 - 6x + 9$ as a perfect square $(x-3)^2$ allows you to solve it mentally in seconds without writing down the intermediate steps. ### Common Pitfall A frequent error is forgetting that $10^0 = 1$ and mistakenly assuming $10^0 = 0$, leading to the equation $x^2 - 6x + 10 = 0$, which has complex roots and will derail the entire solving process. ### Final Answer **Therefore, the correct answer is 3.**