Solve the equation $\log_{3}(x-8)=2-\log_{3}x$. Given $\log_{b}a=c$, show that $\log_{1/b}a=-c$.

1. Group the logarithmic terms on one side of the equation

$\log_{3}(x-8) + \log_{3}x = 2$

2. Apply the product law of logarithms

$\log_{3}(x(x-8)) = 2$

3. Convert the logarithmic equation to exponential form

$x(x-8) = 3^2$
$x^2 - 8x = 9$

4. Solve the resulting quadratic equation and check for valid solutions

$x^2 - 8x - 9 = 0$
$(x-9)(x+1) = 0$
$x = 9 \text{ or } x = -1$

Since $x > 8$ for $\log_{3}(x-8)$ to be defined, we reject $x = -1$.

1. Apply the change of base formula to express the logarithm in base $b$.

$\log_{1/b}a = \frac{\log_b a}{\log_b (1/b)}$

2. Simplify the denominator and substitute the given value to complete the proof.

Since $\log_b (1/b) = \log_b (b^{-1}) = -1$,

$\log_{1/b}a = \frac{c}{-1} = -c$