log7(x^2-12x+36)= 0

To solve for x in the equation log7(x^2-12x+36)=0, we need to use the definition of logarithms which states that log(base a) b = c is equivalent to a^c = b.

In this case, we have log7(x^2-12x+36)=0, which means that 7^0 = x^2-12x+36.

Simplifying 7^0, we get 1, so x^2-12x+36 = 1.

Rearranging the equation, we get x^2-12x+35 = 0.

We can factor this equation as (x-5)(x-7) = 0.

Therefore, the solutions are x = 5 and x = 7.

Checking these solutions, we can see that log7(1) = 0, so both x = 5 and x = 7 satisfy the original equation.