Simplify and test your answers for the following exponential expressions: c2 c4, m4 m5, p7 p3, q3 q7, x5 * x8.

$$\text{Original: } 2^2 \times 2^4 = 4 \times 16 = 64$$ $$\text{Simplified: } 2^6 = 64$$

The simplified expression is $\boxed{c^6}$.

2.b) $m^4 \times m^5 =$

$$\text{Original: } 3^4 \times 3^5 = 81 \times 243 = 19683$$ $$\text{Simplified: } 3^9 = 19683$$

The simplified expression is $\boxed{m^9}$.

2.c) $p^7 \times p^3 =$

$$\text{Original: } 2^7 \times 2^3 = 128 \times 8 = 1024$$ $$\text{Simplified: } 2^{10} = 1024$$

The simplified expression is $\boxed{p^{10}}$.

2.d) $q^3 \times q^7 =$

$$\text{Original: } 3^3 \times 3^7 = 27 \times 2187 = 59049$$ $$\text{Simplified: } 3^{10} = 59049$$

The simplified expression is $\boxed{q^{10}}$.

2.e) $x^5 \times x^8 =$

$$\text{Original: } 4^5 \times 4^8 = 1024 \times 65536 = 67108864$$ $$\text{Simplified: } 4^{13} = 67108864$$

The simplified expression is $\boxed{x^{13}}$.

2.f) $s^9 \times s^2 =$

$$\text{Original: } 5^9 \times 5^2 = 1953125 \times 25 = 48828125$$ $$\text{Simplified: } 5^{11} = 48828125$$

The simplified expression is $\boxed{s^{11}}$.

3. Why can we say: $a^m \times a^n = a^{m+n}$? Give three examples.

We can say $a^m \times a^n = a^{m+n}$ because when you multiply exponential terms that have the same base, you are essentially counting the total number of times that base is multiplied by itself. The exponent $m$ indicates $a$ is multiplied $m$ times, and $n$ indicates $a$ is multiplied $n$ times. When you multiply these two expressions, you combine all those multiplications, resulting in $a$ being multiplied a total of $m+n$ times.

Here are three examples:

a.

$$2^3 \times 2^2$$

This means $(2 \times 2 \times 2) \times (2 \times 2)$.
Counting all the 2s being multiplied, we have $2 \times 2 \times 2 \times 2 \times 2$, which is $2^5$.
Using the rule, $2^{3+2} = 2^5$.

b.

$$x^4 \times x^1$$

This means $(x \times x \times x \times x) \times (x)$.
Counting all the $x$s being multiplied, we have $x \times x \times x \times x \times x$, which is $x^5$.
Using the rule, $x^{4+1} = x^5$.

c.

$$y^2 \times y^3$$

This means $(y \times y) \times (y \times y \times y)$.
Counting all the $y$s being multiplied, we have $y \times y \times y \times y \times y$, which is $y^5$.
Using the rule, $y^{2+3} = y^5$.