The roots of the equation $\log_{2}(x^{2}-4x+5)=(x-2)$ are

Convert the logarithmic equation into its exponential form to get x squared minus 4x plus 5 equals 2 raised to the power of (x minus 2). Adding 4 to both sides gives x squared minus 4x plus 9 equals 4 plus 2 raised to the power of (x minus 2), which can be written as (x minus 3) squared plus 4 equals 4 plus 2 raised to the power of (x minus 3). Testing x equals 3 makes the left side equal to 4 and the right side equal to 2 to the power of 0 plus 4, which is 5, so this approach must be evaluated directly at the options. Substituting x equals 2 into the original equation yields the base of the logarithm as 2 squared minus 8 plus 5, which equals 1, and the logarithm of 1 to any base is 0, perfectly matching the right side (2 minus 2). Substituting x equals 3 yields the base as 3 squared minus 12 plus 5, which equals 2, and log base 2 of 2 is 1, exactly matching the right side (3 minus 2). The valid roots are 2 and 3.

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