Exponential Expressions: Multiplying, Dividing, and Power Rules ... - Free Printable

Let's solve each section of this exponential expressions worksheet step by step, applying the laws of exponents:

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🔷 Exponent Rules Recap:

1. Power of a Power: $(a^m)^n = a^{m \cdot n}$
2. Product of Powers: $a^m \cdot a^n = a^{m+n}$
3. Quotient of Powers: $\frac{a^m}{a^n} = a^{m-n}$
4. Negative Exponent: $a^{-n} = \frac{1}{a^n}$ or $\frac{1}{a^{-n}} = a^n$
5. Power of a Product: $(ab)^n = a^n b^n$
6. Zero Exponent: $a^0 = 1$ (if $a \neq 0$)

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## ✔ Section A: Simplify the following

1) $(x^3)^2$

Use power of a power:
$$
(x^3)^2 = x^{3 \cdot 2} = x^6
$$

2) $(2x)^3$

Apply power to both coefficient and variable:
$$
(2x)^3 = 2^3 \cdot x^3 = 8x^3
$$

3) $(9x^2y)^2$

Apply power to each part:
$$
(9x^2y)^2 = 9^2 \cdot (x^2)^2 \cdot y^2 = 81 \cdot x^4 \cdot y^2 = 81x^4y^2
$$

4) $(5x^2y^4)^2$

$$
= 5^2 \cdot (x^2)^2 \cdot (y^4)^2 = 25 \cdot x^4 \cdot y^8 = 25x^4y^8
$$

5) $(3xy^{-8})^3$

$$
= 3^3 \cdot x^3 \cdot (y^{-8})^3 = 27 \cdot x^3 \cdot y^{-24} = 27x^3y^{-24}
$$

6) $(4x^{-6}y^{-7})^3$

$$
= 4^3 \cdot (x^{-6})^3 \cdot (y^{-7})^3 = 64 \cdot x^{-18} \cdot y^{-21} = 64x^{-18}y^{-21}
$$

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## ✔ Section B: Simplify the following

1) $a^3 \times a^5$

Add exponents:
$$
a^{3+5} = a^8
$$

2) $h \times h^5$

$h = h^1$, so:
$$
h^{1+5} = h^6
$$

3) $n^3 \div n^9$

Subtract exponents:
$$
n^{3-9} = n^{-6}
$$

4) $b^6 \div b^{-14}$

$$
b^{6 - (-14)} = b^{6+14} = b^{20}
$$

5) $4x^{12} \div 2x^3$

Divide coefficients and subtract exponents:
$$
\frac{4}{2} \cdot x^{12-3} = 2x^9
$$

6) $12k^{-5} \times 7k^{-9}$

Multiply coefficients and add exponents:
$$
(12 \cdot 7) \cdot k^{-5 + (-9)} = 84k^{-14}
$$

7) $3a^{-1} \times 3a^{-7}$

$$
(3 \cdot 3) \cdot a^{-1 + (-7)} = 9a^{-8}
$$

8) $48d^{-13} \div 6d^{-18}$

$$
\frac{48}{6} \cdot d^{-13 - (-18)} = 8 \cdot d^{5} = 8d^5
$$

9) $a^2b^6 \times a^4b^8$

Group like bases:
$$
a^{2+4} \cdot b^{6+8} = a^6b^{14}
$$

10) $z^{-4}y^{-1} \times z^{-7}y^{14}$

$$
z^{-4 + (-7)} \cdot y^{-1 + 14} = z^{-11}y^{13}
$$

11) $4q^8 \times 5r \times 2q^{-5} \times 2r^{-4}$

Multiply all coefficients:
$4 \cdot 5 \cdot 2 \cdot 2 = 80$

Now combine variables:
- $q^{8 + (-5)} = q^3$
- $r^{1 + (-4)} = r^{-3}$

So:
$$
80q^3r^{-3} = \frac{80q^3}{r^3}
$$

12) $24s^{-3}t \div 4s^{-3}t^{-8}$

Divide coefficients: $\frac{24}{4} = 6$

For $s$: $s^{-3 - (-3)} = s^0 = 1$

For $t$: $t^{1 - (-8)} = t^{9}$

So:
$$
6 \cdot 1 \cdot t^9 = 6t^9
$$

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## ✔ Section C: Simplify the following

1) $\frac{48a^3}{12a^8}$

Simplify coefficients: $\frac{48}{12} = 4$

Exponent: $a^{3-8} = a^{-5}$

$$
= 4a^{-5} = \frac{4}{a^5}
$$

2) $\frac{30b^{-7}}{5b^5}$

$\frac{30}{5} = 6$, $b^{-7 - 5} = b^{-12}$

$$
= 6b^{-12} = \frac{6}{b^{12}}
$$

3) $\frac{18e^9}{6e^{-3}}$

$\frac{18}{6} = 3$, $e^{9 - (-3)} = e^{12}$

$$
= 3e^{12}
$$

4) $\frac{x^{13}y^2}{x^5y^{-7}}$

$x^{13-5} = x^8$, $y^{2 - (-7)} = y^9$

$$
= x^8y^9
$$

5) $\frac{7b^5 \times 6b}{3b^3}$

First multiply numerator: $7 \cdot 6 = 42$, $b^5 \cdot b = b^6$

So: $\frac{42b^6}{3b^3} = \frac{42}{3} \cdot b^{6-3} = 14b^3$

6) $\frac{12c^{-4} \times 2c^5}{6c^{-2}}$

Numerator: $12 \cdot 2 = 24$, $c^{-4 + 5} = c^1 = c$

So: $\frac{24c}{6c^{-2}} = \frac{24}{6} \cdot c^{1 - (-2)} = 4c^3$

7) $\frac{20d^{-8}}{5d^2 \times 2d^{-9}}$

Denominator: $5 \cdot 2 = 10$, $d^{2 + (-9)} = d^{-7}$

So: $\frac{20d^{-8}}{10d^{-7}} = \frac{20}{10} \cdot d^{-8 - (-7)} = 2 \cdot d^{-1} = \frac{2}{d}$

8) $\frac{24x^{-2}}{3x^{-4} \times 2x^{-7}}$

Denominator: $3 \cdot 2 = 6$, $x^{-4 + (-7)} = x^{-11}$

So: $\frac{24x^{-2}}{6x^{-11}} = \frac{24}{6} \cdot x^{-2 - (-11)} = 4x^9$

9) $\frac{12k^{-16} \times 5m^{-5}}{2m^{-2} \times 15k^4}$

Numerator: $12 \cdot 5 = 60$, $k^{-16}$, $m^{-5}$

Denominator: $2 \cdot 15 = 30$, $m^{-2}$, $k^4$

So:
$\frac{60}{30} = 2$

For $k$: $k^{-16 - 4} = k^{-20}$

For $m$: $m^{-5 - (-2)} = m^{-3}$

Result:
$$
2k^{-20}m^{-3} = \frac{2}{k^{20}m^3}
$$

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## ✔ Extension Problems

🔹 Left Pyramid: Multiply the two bricks below

We are given a pyramid where each brick is the product of the two bricks directly below it.

Top: $a^6b^{-3}$

Middle left: $a^{-4}b^2$

We need to find the missing middle right brick such that:
$$
a^{-4}b^2 \times (\text{?}) = a^6b^{-3}
$$

Let unknown be $X$. Then:
$$
a^{-4}b^2 \cdot X = a^6b^{-3} \Rightarrow X = \frac{a^6b^{-3}}{a^{-4}b^2} = a^{6 - (-4)}b^{-3 - 2} = a^{10}b^{-5}
$$

So middle right = $a^{10}b^{-5}$

Now bottom row:
Left: $a^3b^2$, Middle: $a^{-7}$, Right: ?

We know:
Middle left = $a^3b^2 \times a^{-7} = a^{3 + (-7)}b^2 = a^{-4}b^2$ → matches!

Now find bottom right such that:
$$
a^{-7} \times (\text{?}) = a^{10}b^{-5}
\Rightarrow ? = \frac{a^{10}b^{-5}}{a^{-7}} = a^{10 - (-7)}b^{-5} = a^{17}b^{-5}
$$

✔ So bottom right brick: $a^{17}b^{-5}$

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🔹 Right Pyramid: Divide the two bricks below (left divided by right)

Top: $a$

Given:
Bottom left: $a^{-6}b^{-1}$
Bottom middle: $a^4b^{-2}$
Bottom right: ?

We are told: top = (left brick) ÷ (right brick)

But we don't know which two bricks are being divided. The instruction says:

> "Divide the two bricks below in this pyramid (the left divided by the right)."

This means:
The top brick is equal to (middle-left) ÷ (middle-right), and similarly for bottom.

But the structure shows:

```
a
/ \
? ?
/ \ / \
a⁻⁶b⁻¹ a⁴b⁻² ?
```

Wait — actually, from the diagram:

It appears that the top brick is formed by dividing the two bricks immediately below it, i.e., the left one divided by the right one.

So:
$$
a = \frac{\text{left}}{\text{right}} = \frac{a^{-6}b^{-1}}{a^4b^{-2}} \quad \text{(but wait, that would give } a^{-10}b^{1}, \text{ not } a)
$$

Wait — maybe the middle layer is formed from the bottom layer?

Let’s assume the pyramid works as:

- Each upper brick is the result of the operation on the two below it.
- For the division pyramid, each brick above is left divided by right.

So:

Top: $a$

Below it: two bricks, say $A$ and $B$, such that:
$$
a = \frac{A}{B}
$$

Then $A$ and $B$ are made from the bottom row.

Let’s denote:

Bottom row:
- Left: $a^{-6}b^{-1}$
- Middle: $a^4b^{-2}$
- Right: $?$

Middle layer:
- Left: $a^{-6}b^{-1} \div a^4b^{-2}$? But no — division is left divided by right, so likely:

Each brick in the next level up is formed by dividing the two bricks directly below it.

So:

- Middle-left brick = (bottom-left) ÷ (bottom-middle) = $\frac{a^{-6}b^{-1}}{a^4b^{-2}} = a^{-10}b^{1} = a^{-10}b$

- Middle-right brick = (bottom-middle) ÷ (bottom-right) = $\frac{a^4b^{-2}}{?}$

Then top = (middle-left) ÷ (middle-right) = $a$

So:
$$
\frac{a^{-10}b}{\frac{a^4b^{-2}}{?}} = a
\Rightarrow a^{-10}b \cdot \frac{?}{a^4b^{-2}} = a
\Rightarrow ? \cdot a^{-14}b^{3} = a
\Rightarrow ? = \frac{a}{a^{-14}b^3} = a^{15}b^{-3}
$$

Alternatively, perhaps the top is simply the result of dividing the two middle bricks, and the middle bricks are products?

But the problem says: “Divide the two bricks below in this pyramid (the left divided by the right).”

And the top is $a$, so:

Let’s suppose:

- The two bricks just below the top are: $X$ and $Y$, and $a = \frac{X}{Y}$

But only one of them is shown: the left one is $a^{-6}b^{-1}$, and the right one is $a^4b^{-2}$?

Wait — looking at the image:

The bottom row has three bricks:
- Left: $a^{-6}b^{-1}$
- Middle: $a^4b^{-2}$
- Right: ???

Then the middle row has two bricks (above the bottom row):

- One above left and middle
- One above middle and right

And top is above those two.

But the instruction says: “Divide the two bricks below in this pyramid (the left divided by the right).”

So the top brick is formed by taking the left brick (of the two below) divided by the right brick.

So:

Let the two middle bricks be $A$ and $B$, then:
$$
a = \frac{A}{B}
$$

Now, how are $A$ and $B$ formed?

Likely:

- $A$ = (bottom-left) × (bottom-middle) → but the operation isn't specified.

Wait — the left pyramid was multiplication, so probably this one is division.

But the operation is consistent per pyramid.

Looking back:

- First pyramid: multiply the two bricks below to get the one above.
- Second pyramid: divide the two bricks below (left divided by right).

So for the second pyramid:

Each upper brick is obtained by dividing the left brick by the right brick of the lower level.

So:

Start from bottom:

Bottom row:
- $a^{-6}b^{-1}$, $a^4b^{-2}$, $?$

We need to fill in the blanks so that the top is $a$.

Let’s work upward.

Let’s define:

- Middle-left brick = (bottom-left) ÷ (bottom-middle) = $\frac{a^{-6}b^{-1}}{a^4b^{-2}} = a^{-10}b^{1} = a^{-10}b$

- Middle-right brick = (bottom-middle) ÷ (bottom-right) = $\frac{a^4b^{-2}}{?}$

Then top = (middle-left) ÷ (middle-right) = $a$

So:
$$
\frac{a^{-10}b}{\frac{a^4b^{-2}}{?}} = a
\Rightarrow a^{-10}b \cdot \frac{?}{a^4b^{-2}} = a
\Rightarrow ? \cdot a^{-14}b^{3} = a
\Rightarrow ? = \frac{a}{a^{-14}b^3} = a^{15}b^{-3}
$$

So the bottom-right brick is $a^{15}b^{-3}$

Now, check if this makes sense.

But wait — we have only one unknown, so this should work.

Thus, final answer:

Bottom-right brick: $a^{15}b^{-3}$

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## ✔ Final Answers Summary

Section A

1. $x^6$
2. $8x^3$
3. $81x^4y^2$
4. $25x^4y^8$
5. $27x^3y^{-24}$
6. $64x^{-18}y^{-21}$

Section B

1. $a^8$
2. $h^6$
3. $n^{-6}$
4. $b^{20}$
5. $2x^9$
6. $84k^{-14}$
7. $9a^{-8}$
8. $8d^5$
9. $a^6b^{14}$
10. $z^{-11}y^{13}$
11. $80q^3r^{-3}$ or $\frac{80q^3}{r^3}$
12. $6t^9$

Section C

1. $4a^{-5}$ or $\frac{4}{a^5}$
2. $6b^{-12}$ or $\frac{6}{b^{12}}$
3. $3e^{12}$
4. $x^8y^9$
5. $14b^3$
6. $4c^3$
7. $\frac{2}{d}$
8. $4x^9$
9. $\frac{2}{k^{20}m^3}$

Extension

#### Left Pyramid (Multiplication):
- Missing middle-right: $a^{10}b^{-5}$
- Missing bottom-right: $a^{17}b^{-5}$

#### Right Pyramid (Division):
- Missing bottom-right: $a^{15}b^{-3}$

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