Log_2(x-2)+log_2(x-3)=1

To solve the given logarithmic equation \( \log_{2}(x-2)+\log_{2}(x-3)=1 \), we will first combine the logarithms using the logarithmic rule: \( \log_{a}M + \log_{a}N = \log_{a}(M \cdot N) \).

So, applying this rule we get:

\[ \log_{2}((x-2)(x-3)) = 1 \]

Now, we can rewrite the equation in exponential form using the definition of logarithm:

\[ 2^{1} = (x-2)(x-3) \]

Simplify:

\[ 2 = x^2 - 5x + 6 \]

Rearrange and simplify further:

\[ x^2 - 5x + 4 = 0 \]

Now, we need to factor the quadratic equation to solve for x:

\[ (x-4)(x-1) = 0 \]

Solving for x, we get two possible values for x:

\[ x = 4 \]

or

\[ x = 1 \]

Therefore, the solutions to the given equation are \( x = 4 \) or \( x = 1 \).