If `5^(-5y)=(1)/(3125) and 9^(x)=(1)/(81)`, then `x-y=` _____

Options

A. `-5`
B. `-2`
C. `-3`
D. 2

Answer

To solve the equations \(5^{-5y} = \frac{1}{3125}\) and \(9^{x} = \frac{1}{81}\), we will find the values of \(x\) and \(y\) and then calculate \(x - y\).

### Step 1: Solve for \(y\) from the first equation

We start with the equation:
\[ 5^{-5y} = \frac{1}{3125} \]

Next, we need to express \(3125\) as a power of \(5\). We can find the prime factorization of \(3125\):
\[ 3125 = 5 \times 625 = 5 \times 5 \times 125 = 5 \times 5 \times 5 \times 25 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5 \]

Thus, we can rewrite the equation as:
\[ 5^{-5y} = \frac{1}{5^5} \]

Using the property of exponents that states \(\frac{1}{a^m} = a^{-m}\), we can write:
\[ \frac{1}{5^5} = 5^{-5} \]

Now, we have:
\[ 5^{-5y} = 5^{-5} \]

Since the bases are the same, we can set the exponents equal to each other:
\[ -5y = -5 \]

Dividing both sides by \(-5\):
\[ y = 1 \]

### Step 2: Solve for \(x\) from the second equation

Now we look at the second equation:
\[ 9^{x} = \frac{1}{81} \]

We can express \(81\) as a power of \(9\):
\[ 81 = 9^2 \]

Thus, we can rewrite the equation as:
\[ 9^{x} = \frac{1}{9^2} \]

Using the same property of exponents:
\[ \frac{1}{9^2} = 9^{-2} \]

Now, we have:
\[ 9^{x} = 9^{-2} \]

Again, since the bases are the same, we can set the exponents equal to each other:
\[ x = -2 \]

### Step 3: Calculate \(x - y\)

Now that we have the values of \(x\) and \(y\):
\[ x = -2 \quad \text{and} \quad y = 1 \]

We can find \(x - y\):
\[ x - y = -2 - 1 = -3 \]

Thus, the final answer is:
\[ \boxed{-3} \]