### Instruction: If $\log_9 (x-2)=\frac{1}{2}$, find $\log_{625} x$.### Response: This problem involves logarithms, which are the inverse functions of exponentials.
That means I can rewrite the first equation in exponential form, using the base 9 and the exponent $\frac{1}{2}$.
So, $\log_9 (x-2)=\frac{1}{2}$ is equivalent to $9^{\frac{1}{2}}=x-2$.
I can simplify $9^{\frac{1}{2}}$ by taking the square root of 9, which is 3.
So, I get $3=x-2$.
Adding 2 to both sides, I find that $x=5$.
Now, I can use this value of x to find $\log_{625} x$.
Again, I can use the exponential form, with base 625 and exponent $\log_{625} x$.
So, $\log_{625} x$ is equivalent to $625^{\log_{625} x}=x$.
Since $x=5$, I can plug that in and get $625^{\log_{625} x}=5$.
Now, I need to find what exponent of 625 gives me 5.
One way to do that is to use the change of base formula for logarithms, which says that $\log_b a = \frac{\log_c a}{\log_c b}$, where c is any positive base.
I can choose c to be any convenient base, such as 10, and apply the formula to both sides of the equation.
I get $\log_{10} (625^{\log_{625} x})=\log_{10} 5$.
Using the power rule of logarithms, which says that $\log_b (a^p) = p \log_b a$, I can bring down the exponent on the left side.
I get $(\log_{625} x) \log_{10} 625 = \log_{10} 5$.
I can use a calculator to find the values of $\log_{10} 625$ and $\log_{10} 5$.
They are approximately 2.79588 and 0.69897, respectively.
So, I get $(\log_{625} x) (2.79588) = 0.69897$.
Dividing both sides by 2.79588, I find that $\log_{625} x \approx 0.25001$.
This is the final answer.