Solve the Exponent Equation: 45^-80 × 1/45^-81 × 49 × 7^-5

To solve the problem, let's follow these steps:

• Step 1: Simplify the expression $45^{-80} \cdot \frac{1}{45^{-81}}$.
• Step 2: Simplify $49 \cdot 7^{-5}$.
• Step 3: Combine results to get the final expression.

Now, let's work through each step:
Step 1: Simplify $45^{-80} \cdot \frac{1}{45^{-81}}$. Using the exponent rule: $a^{-b} = \frac{1}{a^b}$ and $a^m \cdot a^n = a^{m+n}$, we have:

$45^{-80} \cdot \frac{1}{45^{-81}} = 45^{-80} \cdot 45^{81} = 45^{-80 + 81} = 45^1 = 45.$

Step 2: Simplify $49 \cdot 7^{-5}$. Note that $49 = 7^2$, so we can rewrite this as: $49 \cdot 7^{-5} = 7^2 \cdot 7^{-5} = 7^{2-5} = 7^{-3} = \frac{1}{7^3}.$

Step 3: Combine these results: $45 \cdot \frac{1}{7^3} = \frac{45}{7^3}.$

Therefore, the solution to the problem is $\frac{45}{7^3}$.