Question

properties of exponents
#1 \\(\frac{9^2}{9^7}\\)
#2 \\((5^6)(5^3)\\)
#3 \\((7^5)^6\\)
#4 \\((x^5)^3\\)
#5 \\(\frac{a^5}{a^5}\\)
#6 \\((7^{-7})(7^{12})\\)
#7 \\(\frac{8^4}{8^5} \cdot 8^9\\)
#8 \\((3^5 \cdot 3^2)^6\\)
#9 \\(\frac{10^7}{10^5} \cdot 10^{-5}\\)
#10 \\((5^3)(5^4)(5)\\)
#11 \\(\frac{x^{11}}{x^4 \cdot x^6}\\)
#12 \\((a^{-4})^2(a^5)^4\\)
#13 \\(\frac{a^7}{a} \cdot a^4\\)
#14 \\(\frac{7^{-3} \cdot 7^8}{7^4 \cdot 7^{-5}}\\)
#15 \\((x^3)^4(x^3)^{-2}\\)
#16 \\((4^3)(4^{-3})\\)
#17 \\((9^2)^5(9^2)(9^4)\\)
#18 \\((5^2)^6(5^{-13})(5^7)\\)
#19 \\(\frac{17^7}{17^5} \cdot (17^{-5})^2\\)
#20 \\(\frac{x \cdot x^5}{x^3 \cdot x^6}\\)

Explanation:

Let's solve problem #1: $\boldsymbol{\frac{9^2}{9^7}}$

Step1: Recall the quotient rule for exponents.

The quotient rule states that $\frac{a^m}{a^n} = a^{m - n}$ when $a
eq 0$. Here, $a = 9$, $m = 2$, and $n = 7$.

Step2: Apply the quotient rule.

Substitute into the formula: $\frac{9^2}{9^7} = 9^{2 - 7}$.

Step3: Simplify the exponent.

Calculate $2 - 7 = -5$, so we have $9^{-5}$.

Step4: Rewrite with positive exponent (optional, but standard).

Using the definition $a^{-n} = \frac{1}{a^n}$, we get $9^{-5} = \frac{1}{9^5}$. If we calculate $9^5 = 59049$, then $\frac{1}{9^5} = \frac{1}{59049}$. However, leaving it as $9^{-5}$ or $\frac{1}{9^5}$ is also correct. For simplicity, we’ll use the exponent form first.

Answer:

$\boldsymbol{9^{-5}}$ (or $\boldsymbol{\frac{1}{59049}}$)