Log2(x-2)+log2(x-3)=1

To solve the given logarithmic equation, we can combine the two logarithms using the properties of logarithms.

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

Using the product rule of logarithms, we can combine the two logarithms into a single logarithm:

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

Now, we can rewrite the equation in exponential form:

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

\( 2 = x^{2} - 5x + 6 \)

Rearranging the equation and setting it equal to zero:

\( x^{2} - 5x + 4 = 0 \)

Now, we can solve the quadratic equation by factoring or using the quadratic formula:

\( x^{2} - 4x - x + 4 = 0 \)

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

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

This gives us two solutions:

\( x = 4 \) or \( x = 1 \)

Therefore, the possible values for x that satisfy the given logarithmic equation are x = 4 and x = 1.