To solve the given logarithmic equation, we can start by combining the two logarithms using the product rule of logarithms:
log2[(x-6)(x-8)] = 3
Next, we will use the fact that logarithmic equations can be rewritten in exponential form. In this case, we rewrite the equation as:
2^3 = (x-6)(x-8)
Solving for x, we get:
8 = x^2 - 14x + 48
Rearranging terms:
x^2 - 14x + 40 = 0
This is a quadratic equation that can be factored as:
(x-10)(x-4) = 0
Setting each factor to 0:
x - 10 = 0 => x = 10x - 4 = 0 => x = 4
Therefore, the solutions to the logarithmic equation are x = 10 and x = 4.
To solve the given logarithmic equation, we can start by combining the two logarithms using the product rule of logarithms:
log2[(x-6)(x-8)] = 3
Next, we will use the fact that logarithmic equations can be rewritten in exponential form. In this case, we rewrite the equation as:
2^3 = (x-6)(x-8)
Solving for x, we get:
8 = x^2 - 14x + 48
Rearranging terms:
x^2 - 14x + 40 = 0
This is a quadratic equation that can be factored as:
(x-10)(x-4) = 0
Setting each factor to 0:
x - 10 = 0 => x = 10
x - 4 = 0 => x = 4
Therefore, the solutions to the logarithmic equation are x = 10 and x = 4.