Exponential Form Worksheet: Match each exponential form with its expanded form.

Explanation:
We are given a list of exponential forms (like $5^3$, $2^5$, etc.) and a list of expanded forms (like $11 \cdot 11 \cdot 11$, $17 \cdot 17 \cdot 17 \cdot 17$, etc.). Our job is to match each exponential form with its correct expanded form.

Recall:
- An expression like $a^n$ means multiply $a$ by itself $n$ times.
- So $5^3 = 5 \cdot 5 \cdot 5$, not $5 \cdot 3$ or anything else.

Let’s go one by one:

1. $5^3$: base = 5, exponent = 3 → three 5s multiplied: $5 \cdot 5 \cdot 5$.
Look at the right column — do we see that? Yes: 5 · 5 · 5 is listed. So match $5^3$ ↔ $5 \cdot 5 \cdot 5$

2. $2^5$: base = 2, exponent = 5 → five 2s: $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$.
Right column has: 2 · 2 · 2 · 2 · 2 — yes! Match $2^5$ ↔ $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$

3. $7^4$: base = 7, exponent = 4 → four 7s: $7 \cdot 7 \cdot 7 \cdot 7$.
Right column has: 7 · 7 · 7 · 7 — match $7^4$ ↔ $7 \cdot 7 \cdot 7 \cdot 7$

4. $11^3$: base = 11, exponent = 3 → three 11s: $11 \cdot 11 \cdot 11$.
Right column has: 11 · 11 · 11 — match $11^3$ ↔ $11 \cdot 11 \cdot 11$

5. $3^5$: base = 3, exponent = 5 → five 3s: $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$.
Right column has: 3 · 3 · 3 · 3 · 3 — match $3^5$ ↔ $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$

6. $5^4$: base = 5, exponent = 4 → four 5s: $5 \cdot 5 \cdot 5 \cdot 5$.
Wait — look carefully: Is there a “5 · 5 · 5 · 5”? No. The right column has:
- 11·11·11
- 17·17·17·17
- 2·2·2·2·2
- 5·5·5
- 2·2·2·2
- 7·7·7·7
- 3·3·3·3·3
- 2·2·2·2·2 (already used)
- 17⁴ is also there — wait, maybe we misread.

Let me list all left-side expressions clearly:

Left column:
- $5^3$
- $2^5$
- $7^4$
- $11^3$
- $3^5$
- $5^4$
- $2^6$
- $17^4$
- $3^4$
- $2^4$

Right column (10 items):
- 11 · 11 · 11
- 17 · 17 · 17 · 17
- 2 · 2 · 2 · 2 · 2
- 5 · 5 · 5
- 2 · 2 · 2 · 2
- 7 · 7 · 7 · 7
- 3 · 3 · 3 · 3 · 3
- 2 · 2 · 2 · 2 · 2 ← duplicate? Wait, no — let me count again.

Actually, looking back at the image description (text provided), the right column is:

• 11 . 11 . 11
• 17 . 17 . 17 . 17
• 2 . 2 . 2 . 2 . 2
• 5 . 5 . 5
• 2 . 2 . 2 . 2
• 7 . 7 . 7 . 7
• 3 . 3 . 3 . 3 . 3
• 2 . 2 . 2 . 2 . 2 ← this appears again? No — wait, original text says:

"• 11 . 11 . 11
• 17 . 17 . 17 . 17
• 2 . 2 . 2 . 2 . 2
• 5 . 5 . 5
• 2 . 2 . 2 . 2
• 7 . 7 . 7 . 7
• 3 . 3 . 3 . 3 . 3
• 2 . 2 . 2 . 2 . 2
• 5 . 5 . 5 . 5
• 3 . 3 . 3 . 3"

Wait — in the user-provided text, it's:

> • 11 . 11 . 11
> • 17 . 17 . 17 . 17
> • 2 . 2 . 2 . 2 . 2
> • 5 . 5 . 5
> • 2 . 2 . 2 . 2
> • 7 . 7 . 7 . 7
> • 3 . 3 . 3 . 3 . 3
> • 2 . 2 . 2 . 2 . 2
> • 5 . 5 . 5 . 5
> • 3 . 3 . 3 . 3

But that’s 10 items — matches 10 left items.

Let me rewrite both sides clearly:

Left (exponential forms):
1. $5^3$
2. $2^5$
3. $7^4$
4. $11^3$
5. $3^5$
6. $5^4$
7. $2^6$
8. $17^4$
9. $3^4$
10. $2^4$

Right (expanded forms):
A. $11 \cdot 11 \cdot 11$ → 3 elevens → $11^3$
B. $17 \cdot 17 \cdot 17 \cdot 17$ → 4 seventeens → $17^4$
C. $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ → 5 twos → $2^5$
D. $5 \cdot 5 \cdot 5$ → 3 fives → $5^3$
E. $2 \cdot 2 \cdot 2 \cdot 2$ → 4 twos → $2^4$
F. $7 \cdot 7 \cdot 7 \cdot 7$ → 4 sevens → $7^4$
G. $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$ → 5 threes → $3^5$
H. $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ → again 5 twos? But we already used that for $2^5$. Hmm — but wait, maybe it's a typo in transcription? Let’s check original string again:

The user wrote:
> • 11 . 11 . 11
> • 17 . 17 . 17 . 17
> • 2 . 2 . 2 . 2 . 2
> • 5 . 5 . 5
> • 2 . 2 . 2 . 2
> • 7 . 7 . 7 . 7
> • 3 . 3 . 3 . 3 . 3
> • 2 . 2 . 2 . 2 . 2
> • 5 . 5 . 5 . 5
> • 3 . 3 . 3 . 3

Ah! There are two entries with “2 · 2 · 2 · 2 · 2” — that can’t be right unless one is actually something else. But notice: the 8th item is “2 . 2 . 2 . 2 . 2”, and the 3rd is also that. However, we have $2^5$ and $2^6$ — $2^6$ needs six 2s, but none of the expanded forms show six 2s.

Wait — maybe the 8th item is actually “2 · 2 · 2 · 2 · 2 · 2”? But the text says “2 . 2 . 2 . 2 . 2” again.

Let me double-check the original problem as given in the prompt — perhaps a formatting issue.

Looking again: the right column as typed:

• 11 . 11 . 11
• 17 . 17 . 17 . 17
• 2 . 2 . 2 . 2 . 2
• 5 . 5 . 5
• 2 . 2 . 2 . 2
• 7 . 7 . 7 . 7
• 3 . 3 . 3 . 3 . 3
• 2 . 2 . 2 . 2 . 2
• 5 . 5 . 5 . 5
• 3 . 3 . 3 . 3

That’s 10 lines. Now count how many factors in each:

1. 11×3 → exponent 3
2. 17×4 → exp 4
3. 2×5 → exp 5
4. 5×3 → exp 3
5. 2×4 → exp 4
6. 7×4 → exp 4
7. 3×5 → exp 5
8. 2×5 → exp 5 again — conflict
9. 5×4 → exp 4
10. 3×4 → exp 4

But left side has exponents:
$5^3$ (3)
$2^5$ (5)
$7^4$ (4)
$11^3$ (3)
$3^5$ (5)
$5^4$ (4)
$2^6$ (6) ← needs 6 twos
$17^4$ (4)
$3^4$ (4)
$2^4$ (4)

We’re missing an expanded form with six 2s. So likely, the 8th item is miswritten — it should be 2 · 2 · 2 · 2 · 2 · 2 (six 2s) for $2^6$.

Given this is a standard worksheet, and $2^6 = 64$, and expanded form must be six 2s, it's almost certain that the eighth item is meant to be six 2s.

Also, notice: there are *two* entries with “2 · 2 · 2 · 2 · 2”, but only one $2^5$. So one of them must be $2^6$.

Let’s assume the intended right column is:

1. 11·11·11
2. 17·17·17·17
3. 2·2·2·2·2 → $2^5$
4. 5·5·5 → $5^3$
5. 2·2·2·2 → $2^4$
6. 7·7·7·7 → $7^4$
7. 3·3·3·3·3 → $3^5$
8. 2·2·2·2·2·2 → $2^6$ *(corrected)*
9. 5·5·5·5 → $5^4$
10. 3·3·3·3 → $3^4$

That gives perfect matching:

- $5^3$ ↔ 5·5·5
- $2^5$ ↔ 2·2·2·2·2
- $7^4$ ↔ 7·7·7·7
- $11^3$ ↔ 11·11·11
- $3^5$ ↔ 3·3·3·3·3
- $5^4$ ↔ 5·5·5·5
- $2^6$ ↔ 2·2·2·2·2·2
- $17^4$ ↔ 17·17·17·17
- $3^4$ ↔ 3·3·3·3
- $2^4$ ↔ 2·2·2·2

Yes — all match uniquely.

So the correct pairings are:

$5^3$ → $5 \cdot 5 \cdot 5$
$2^5$ → $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$
$7^4$ → $7 \cdot 7 \cdot 7 \cdot 7$
$11^3$ → $11 \cdot 11 \cdot 11$
$3^5$ → $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$
$5^4$ → $5 \cdot 5 \cdot 5 \cdot 5$
$2^6$ → $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$
$17^4$ → $17 \cdot 17 \cdot 17 \cdot 17$
$3^4$ → $3 \cdot 3 \cdot 3 \cdot 3$
$2^4$ → $2 \cdot 2 \cdot 2 \cdot 2$

Final Answer:
$5^3$ ↔ $5 \cdot 5 \cdot 5$
$2^5$ ↔ $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$
$7^4$ ↔ $7 \cdot 7 \cdot 7 \cdot 7$
$11^3$ ↔ $11 \cdot 11 \cdot 11$
$3^5$ ↔ $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3$
$5^4$ ↔ $5 \cdot 5 \cdot 5 \cdot 5$
$2^6$ ↔ $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$
$17^4$ ↔ $17 \cdot 17 \cdot 17 \cdot 17$
$3^4$ ↔ $3 \cdot 3 \cdot 3 \cdot 3$
$2^4$ ↔ $2 \cdot 2 \cdot 2 \cdot 2$