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High School Mathematics · Exponent Laws

Factor.
Simplify.
Repeat.

Two patterns. Twenty problems. Master the logic once — every question becomes obvious.

Type A — \(b^{n+k} \pm b^n\) Type B — \(2^{n+p}(3^n - 3^{n+q})\) 20 Questions

Questions 1 – 10

Factor out the base.

Pull out the smallest power. Read off what's left. That's all.

The Pattern

\(b^{n+k} - b^{n+j} = b^n\!\left(b^k - b^j\right)\)
1
Identify the smallest power — that's your \(b^n\)
2
Factor it out from both terms
3
Compute the bracket — that's \(a\)
⚡ SMALLEST POWER → FACTOR → READ OFF a

Common Traps

!
Sign matters: \(b^{n+k} \mathbf{+} b^n = b^n(b^k \mathbf{+} 1)\) — don't subtract!
!
Both terms above \(n\)? Still factor \(b^n\): \(b^{n+3}+b^{n+1}=b^n(b^3+b)\)
!
\(2^3 = 8\) not 6. \(5^2 = 25\) not 20. Check your powers!
No. 01
Type AEasy

For all natural numbers \(n\),  \(3^{n+2} - 3^n = a \times 3^n\).  Find \(a\).

3^n ( 3² − 1 ) → compute the bracket

Step-by-step

1
Factor: \(3^{n+2} - 3^n = 3^n(3^2 - 1)\)
2
Compute: \(3^2 - 1 = 9 - 1 = 8\)

\(a = 8\)

No. 02
Type AEasy

For all natural numbers \(n\),  \(2^{n+3} - 2^n = a \times 2^n\).  Find \(a\).

2^3 = 8, not 6 — common slip!

Step-by-step

1
Factor: \(2^{n+3} - 2^n = 2^n(2^3 - 1)\)
2
Compute: \(8 - 1 = 7\)

\(a = 7\)

No. 03
Type AMedium

For all natural numbers \(n\),  \(5^{n+2} - 5^n = a \times 5^n\).  Find \(a\).

5² = 25, not 20 — don't multiply by 2!

Step-by-step

1
\(5^{n+2} - 5^n = 5^n(5^2 - 1)\)
2
\(25 - 1 = 24\)  ·  So \(a = 24\).

\(a = 24\)

No. 04
Type AMedium

For all natural numbers \(n\),  \(2^{n+4} - 2^{n+1} = a \times 2^n\).  Find \(a\).

Factor 2^n from both — what's left in each term?

Step-by-step

1
\(2^{n+4} - 2^{n+1} = 2^n(2^4 - 2^1)\)
2
\(16 - 2 = 14\)

\(a = 14\)

No. 05
Type AMedium

For all natural numbers \(n\),  \(3^{n+3} - 3^{n+1} = a \times 3^n\).  Find \(a\).

3^n(3³ − 3¹) = 3^n(27 − 3)

Step-by-step

1
\(3^{n+3} - 3^{n+1} = 3^n(3^3 - 3^1)\)
2
\(27 - 3 = 24\)

\(a = 24\)

No. 06
Type AMedium

For all natural numbers \(n\),  \(4^{n+2} - 4^{n+1} = a \times 4^n\).  Find \(a\).

4^n(4² − 4¹) = 4^n(16 − 4)

Step-by-step

1
\(4^{n+2} - 4^{n+1} = 4^n(4^2 - 4^1)\)
2
\(16 - 4 = 12\)

\(a = 12\)

No. 07
Type AMedium

For all natural numbers \(n\),  \(2^{n+5} - 2^{n+2} = a \times 2^n\).  Find \(a\).

2^n(2⁵ − 2²) = 2^n(32 − 4)

Step-by-step

1
\(2^{n+5} - 2^{n+2} = 2^n(2^5 - 2^2)\)
2
\(32 - 4 = 28\)

\(a = 28\)

No. 08
Type AMedium

For all natural numbers \(n\),  \(3^{n+2} + 3^n = a \times 3^n\).  Find \(a\).

PLUS sign → bracket is (9 + 1), not (9 − 1)!

Step-by-step

1
\(3^{n+2} + 3^n = 3^n(3^2 + 1)\)
2
\(9 + 1 = 10\) — the sign is +, not −

\(a = 10\)

No. 09
Type AMedium

For all natural numbers \(n\),  \(2^{n+3} + 2^n = a \times 2^n\).  Find \(a\).

Compare with Q2 — same numbers, but + not −

Step-by-step

1
\(2^{n+3} + 2^n = 2^n(2^3 + 1)\)
2
\(8 + 1 = 9\) — Q2 was \(8-1=7\). One sign changes everything.

\(a = 9\)

No. 10
Type AHard

For all natural numbers \(n\),  \(2^{n+3} + 2^{n+1} = a \times 2^n\).  Find \(a\).

Both exponents above n — factor 2^n from both

Step-by-step

1
\(2^{n+3} + 2^{n+1} = 2^n(2^3 + 2^1)\)
2
\(8 + 2 = 10\)

\(a = 10\)

Questions 11 – 20

Three moves.
Every time.

Split the prefix. Factor the bracket. Combine into \(6^n\). The answer falls out.

The 3-Move Pattern

\(2^{n+p}(3^n - 3^{n+q}) = a \times 6^n\)
①
Split: \(2^{n+p} = 2^p \cdot 2^n\)
②
Factor the bracket: \(3^n(1 - 3^q)\)
③
Combine: \(2^n \cdot 3^n = 6^n\), read off \(a = 2^p(1-3^q)\)
⚡ SPLIT → FACTOR → COMBINE → a

When the bracket starts at 3^(n+s)

s
If bracket is \(3^{n+s} - 3^{n+s+q}\), factor out \(3^{n+s}\):
\(= 3^s \cdot 3^n(1-3^q)\) → extra \(3^s\) multiplies \(a\)
!
Big gap \(q=2\): \((1-3^2) = -8\), not \(-2\). Count the gap carefully.
No. 11
Type BEasy

For all natural numbers \(n\),  \(2^{n+2}(3^n - 3^{n+1}) = a \times 6^n\).  Find \(a\).

The original — 3 moves: SPLIT · FACTOR · COMBINE

Step-by-step

①
Split: \(2^{n+2} = 4 \cdot 2^n\)
②
Factor: \(3^n - 3^{n+1} = 3^n(1-3) = -2 \cdot 3^n\)
③
Combine: \(4 \cdot (-2) \cdot 2^n \cdot 3^n = -8 \cdot 6^n\)

\(a = -8\)

No. 12
Type BEasy

For all natural numbers \(n\),  \(2^{n+3}(3^n - 3^{n+1}) = a \times 6^n\).  Find \(a\).

Same bracket as Q11. Only the prefix changed. Double it?

Step-by-step

①
\(2^{n+3} = 8 \cdot 2^n\)
②
Bracket same: \(-2 \cdot 3^n\)
③
\(8 \cdot (-2) \cdot 6^n = -16 \cdot 6^n\)

\(a = -16\)

No. 13
Type BMedium

For all natural numbers \(n\),  \(2^{n+1}(3^n - 3^{n+2}) = a \times 6^n\).  Find \(a\).

Gap is 2 now → (1 − 3²) = (1 − 9) = −8

Step-by-step

①
\(2^{n+1} = 2 \cdot 2^n\)
②
\(3^n - 3^{n+2} = 3^n(1-3^2) = 3^n \cdot (-8)\)
③
\(2 \cdot (-8) \cdot 6^n = -16 \cdot 6^n\)

\(a = -16\)

No. 14
Type BMedium

For all natural numbers \(n\),  \(2^{n+2}(3^{n+1} - 3^{n+2}) = a \times 6^n\).  Find \(a\).

Bracket starts at 3^(n+1) — extra ×3 hiding inside!

Step-by-step

①
\(2^{n+2} = 4 \cdot 2^n\)
②
\(3^{n+1} - 3^{n+2} = 3^{n+1}(1-3) = -2 \cdot 3 \cdot 3^n = -6 \cdot 3^n\)
③
\(4 \cdot (-6) \cdot 6^n = -24 \cdot 6^n\)

\(a = -24\)

No. 15
Type BHard

For all natural numbers \(n\),  \(2^{n+3}(3^{n+1} - 3^{n+2}) = a \times 6^n\).  Find \(a\).

Large prefix + shifted bracket — two things to track

Step-by-step

①
\(2^{n+3} = 8 \cdot 2^n\)
②
\(3^{n+1} - 3^{n+2} = 3^{n+1}(1-3) = -2 \cdot 3 \cdot 3^n = -6 \cdot 3^n\)
③
\(8 \cdot (-6) \cdot 6^n = -48 \cdot 6^n\)

\(a = -48\)

No. 16
Type BMedium

For all natural numbers \(n\),  \(2^{n+4}(3^n - 3^{n+1}) = a \times 6^n\).  Find \(a\).

2^(n+4) = 16·2^n. Bracket = −2·3^n. Multiply!

Step-by-step

①
\(2^{n+4} = 16 \cdot 2^n\)
②
\(3^n - 3^{n+1} = -2 \cdot 3^n\)
③
\(16 \cdot (-2) \cdot 6^n = -32 \cdot 6^n\)

\(a = -32\)

No. 17
Type BHard

For all natural numbers \(n\),  \(2^n(3^{n+2} - 3^{n+3}) = a \times 6^n\).  Find \(a\).

No prefix split (p=0). Bracket starts at 3^(n+2) → ×9·3^n

Step-by-step

①
\(2^n\) stays as-is (no split)
②
\(3^{n+2} - 3^{n+3} = 3^{n+2}(1-3) = -2 \cdot 9 \cdot 3^n = -18 \cdot 3^n\)
③
\(1 \cdot (-18) \cdot 6^n = -18 \cdot 6^n\)

\(a = -18\)

No. 18
Type BHard

For all natural numbers \(n\),  \(2^{n+2}(3^n - 3^{n+2}) = a \times 6^n\).  Find \(a\).

Gap = 2 → bracket gives (1 − 9) = −8, not −2

Step-by-step

①
\(2^{n+2} = 4 \cdot 2^n\)
②
\(3^n - 3^{n+2} = 3^n(1-3^2) = 3^n(1-9) = -8 \cdot 3^n\)
③
\(4 \cdot (-8) \cdot 6^n = -32 \cdot 6^n\)

\(a = -32\)

No. 19
Type BHard

For all natural numbers \(n\),  \(2^{n+1}(3^{n+1} - 3^{n+2}) = a \times 6^n\).  Find \(a\).

Both prefix AND bracket shifted — track both extra factors

Step-by-step

①
\(2^{n+1} = 2 \cdot 2^n\)
②
\(3^{n+1} - 3^{n+2} = 3^{n+1}(1-3) = -2 \cdot 3 \cdot 3^n = -6 \cdot 3^n\)
③
\(2 \cdot (-6) \cdot 6^n = -12 \cdot 6^n\)

\(a = -12\)

No. 20
Type BHard ★

For all natural numbers \(n\),  \(2^{n+3}(3^n - 3^{n+2}) = a \times 6^n\).  Find \(a\).

Large prefix (×8) AND large gap (÷9) strike together

Step-by-step

①
\(2^{n+3} = 8 \cdot 2^n\)
②
\(3^n - 3^{n+2} = 3^n(1-9) = -8 \cdot 3^n\)
③
\(8 \cdot (-8) \cdot 6^n = -64 \cdot 6^n\)

\(a = -64\)

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