Simplify 7^x × 7^(-x): Multiplying Exponential Expressions

This comes from the pattern of exponents! Look: $7^3 = 343$, $7^2 = 49$, $7^1 = 7$. Each time we decrease the exponent by 1, we divide by 7. So $7^0 = 7^1 ÷ 7 = 1$!

The result would still be 1! For any non-zero number: $5^x \cdot 5^{-x} = 5^0 = 1$ and $10^x \cdot 10^{-x} = 10^0 = 1$. The zero power rule works for all bases.

Yes! Think of $7^{-x}$ as $\frac{1}{7^x}$. Then $7^x \cdot \frac{1}{7^x} = \frac{7^x}{7^x} = 1$. Same result, different approach!

If x = 0, then $7^0 \cdot 7^{-0} = 7^0 \cdot 7^0 = 1 \cdot 1 = 1$. The answer is still 1 regardless of what value x has!

This comes from repeated multiplication! $7^2 \cdot 7^3$ means $(7 \cdot 7) \cdot (7 \cdot 7 \cdot 7)$ which gives us 5 sevens total, so $7^5$. That's why 2 + 3 = 5!