We are given the equation:

$3 \log_{125} (A - 6) = -1$

To solve for $A$, let's go through the following steps:

Step 1: Isolate the logarithm

First, divide both sides by 3 to simplify the equation:

$\log_{125} (A - 6) = \frac{-1}{3}$

Step 2: Convert from logarithmic to exponential form

The logarithmic equation $\log_b(x) = y$ can be rewritten in exponential form as $x = b^y$. Using this property, we rewrite the equation:

$A - 6 = 125^{\frac{-1}{3}}$

Step 3: Simplify the exponential expression

$125^{\frac{1}{3}}$ is the cube root of 125, which equals 5. So:

$125^{\frac{-1}{3}} = \frac{1}{5}$

Now the equation becomes:

$A - 6 = \frac{1}{5}$

Step 4: Solve for $A$

Add 6 to both sides:

$A = 6 + \frac{1}{5}$

Convert 6 to a fraction:

$A = \frac{30}{5} + \frac{1}{5} = \frac{31}{5}$

Final Answer:

$A = \frac{31}{5}$

Would you like more details or have any questions?

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Here are 5 related questions:

1. How do you convert a logarithmic equation to an exponential form?
2. What is the value of $\log_b(x)$ when $x$ is a fraction?
3. How do you simplify fractional exponents?
4. Can logarithms have negative values?
5. How do you add fractions with different denominators?

Tip: When solving logarithmic equations, always check if the argument of the log stays positive in the final solution.