Simplify using a single exponent:

$9^{-8}$
$(-5)^2(-5)^3$
$(-2)^{11}(-2)^9(-2)^3$
$6^7 \cdot 6^9 \cdot 6^3$
$\frac{7^4}{7}$
$\frac{10^2}{10^5}$
$\frac{2^4}{2^4}$
$\frac{(-3)^4}{(-3)^5}$
$\frac{(-7)^5}{(-7) \cdot(-7)^4}$
$\frac{8^2}{8^{-2}}$
$\frac{5^4 \cdot 5^9}{5^3 \cdot 5^3}$
$\frac{4^3 \cdot 4^0}{4^2 \cdot 4^3}$
$\frac{9^{12} \cdot 9^3}{9^7 \cdot 9^8}$
$\frac{m^{-2}}{m^{-5}}$
$\frac{3^3}{3^4}$
$\frac{c^{12}}{c^{15}}$
$(-13)^2(-13)^3$
$(-8)^0(-8)^9(-8)^3$
$(-25)^{11}(-25)^3$

Question

04-11-2026
Mathematics

Answered

Simplify using a single exponent:

$9^{-8}$
$(-5)^2(-5)^3$
$(-2)^{11}(-2)^9(-2)^3$
$6^7 \cdot 6^9 \cdot 6^3$
$\frac{7^4}{7}$
$\frac{10^2}{10^5}$
$\frac{2^4}{2^4}$
$\frac{(-3)^4}{(-3)^5}$
$\frac{(-7)^5}{(-7) \cdot(-7)^4}$
$\frac{8^2}{8^{-2}}$
$\frac{5^4 \cdot 5^9}{5^3 \cdot 5^3}$
$\frac{4^3 \cdot 4^0}{4^2 \cdot 4^3}$
$\frac{9^{12} \cdot 9^3}{9^7 \cdot 9^8}$
$\frac{m^{-2}}{m^{-5}}$
$\frac{3^3}{3^4}$
$\frac{c^{12}}{c^{15}}$
$(-13)^2(-13)^3$
$(-8)^0(-8)^9(-8)^3$
$(-25)^{11}(-25)^3$

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-04T16:03:49-05:00

Simplify expressions using the product rule $a^m \cdot a^n = a^{m+n}$.
Reduce quotients using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$.
Handle zero exponents using the rule $a^0 = 1$.
Calculate final exponent forms:

$9^{-8}$, $(-5)^5$, $(-2)^{23}$, $6^{19}$, $7^3$, $10^{-3}$, $1$, $(-3)^{-1}$, $1$, $8^4$, $5^7$, $4^{-2}$, $1$, $m^3$, $3^{-1}$, $c^{-3}$, $(-13)^5$, $(-8)^{12}$, $(-25)^{14}$ are the simplified forms.

Explanation

Rules of Exponents and Analysis

To simplify expressions involving exponents with the same base, we use the fundamental laws of exponents:

Product Rule: $a^m \cdot a^n = a^{m+n}$
Quotient Rule: $\frac{a^m}{a^n} = a^{m-n}$
Zero Exponent Rule: $a^0 = 1$ (where $a eq 0$)
Single Exponent Rule: $a^1 = a$

We will process each expression by identifying the base and applying the appropriate rule to express it as a single power $a^n$.

Simplifying the Expressions

Here are the simplifications for the provided list:

$9^{-8}$ is already in the form $a^n$.
$(-5)^2 \cdot (-5)^3 = (-5)^{2+3} = (-5)^5$.
$(-2)^{11} \cdot (-2)^9 \cdot (-2)^3 = (-2)^{11+9+3} = (-2)^{23}$.
$6^7 \cdot 6^9 \cdot 6^3 = 6^{7+9+3} = 6^{19}$.
$\frac{7^4}{7^1} = 7^{4-1} = 7^3$.
$\frac{10^2}{10^5} = 10^{2-5} = 10^{-3}$.
$\frac{2^4}{2^4} = 2^{4-4} = 2^0 = 1$ (Note: 1 can be written as $a^0$ for any base $a eq 0$).
$\frac{(-3)^4}{(-3)^5} = (-3)^{4-5} = (-3)^{-1}$.
$\frac{(-7)^5}{(-7)^1 \cdot (-7)^4} = \frac{(-7)^5}{(-7)^{1+4}} = \frac{(-7)^5}{(-7)^5} = (-7)^0 = 1$.
$\frac{8^2}{8^{-2}} = 8^{2-(-2)} = 8^{2+2} = 8^4$.
$\frac{5^4 \cdot 5^9}{5^3 \cdot 5^3} = \frac{5^{13}}{5^6} = 5^{13-6} = 5^7$.
$\frac{4^3 \cdot 4^0}{4^2 \cdot 4^3} = \frac{4^{3+0}}{4^{2+3}} = \frac{4^3}{4^5} = 4^{3-5} = 4^{-2}$.
$\frac{9^{12} \cdot 9^3}{9^7 \cdot 9^8} = \frac{9^{15}}{9^{15}} = 9^0 = 1$.
$\frac{m^{-2}}{m^{-5}} = m^{-2-(-5)} = m^{-2+5} = m^3$.
$\frac{3^3}{3^4} = 3^{3-4} = 3^{-1}$.
$\frac{c^{12}}{c^{15}} = c^{12-15} = c^{-3}$.
$(-13)^2 \cdot (-13)^3 = (-13)^{2+3} = (-13)^5$.
$(-8)^0 \cdot (-8)^9 \cdot (-8)^3 = 1 \cdot (-8)^{12} = (-8)^{12}$.
$(-25)^{11} \cdot (-25)^3 = (-25)^{11+3} = (-25)^{14}$.

Final Summary of Results

The simplification process demonstrates how the laws of exponents efficiently reduce complex expressions to a single power. Each expression was resolved by either summing exponents (for products) or subtracting them (for quotients). The final results for each are:

$9^{-8}$
$(-5)^5$
$(-2)^{23}$
$6^{19}$
$7^3$
$10^{-3}$
$2^0=1$
$(-3)^{-1}$
$(-7)^0=1$
$8^4$
$5^7$
$4^{-2}$
$9^0=1$
$m^3$
$3^{-1}$
$c^{-3}$
$(-13)^5$
$(-8)^{12}$
$(-25)^{14}$

Examples

Understanding exponents is crucial in fields like computer science and finance. For instance, if you are calculating the growth of a server's data capacity where each day the number of users triples ($3^1$, $3^2$, $3^3$, etc.), or if you are calculating compound interest where an investment grows exponentially over time, the laws of exponents allow you to easily calculate the total magnitude of growth without performing tedious multiplications. This fundamental skill forms the backbone of managing exponential systems in engineering, biology, and economics.