You open a problem and stare at this:

\frac{2^{10} \cdot 3^{10}}{6^{10}}.

The first instinct is to panic. 2^{10} = 1024. 3^{10} = 59{,}049. 6^{10} = 60{,}466{,}176. Even on a calculator, you would have to multiply 1024 \cdot 59{,}049 (which is 60{,}466{,}176) and then divide by 60{,}466{,}176. The answer is 1, but the arithmetic is brutal.

There is a one-line shortcut. Spot the pattern — two powers with the same exponent — and regroup.

\frac{2^{10} \cdot 3^{10}}{6^{10}} = \frac{(2 \cdot 3)^{10}}{6^{10}} = \frac{6^{10}}{6^{10}} = 1.

Three steps, no arithmetic. Every step is a single application of an exponent law you already know.

The key recognition — a^n \cdot b^n = (ab)^n

The entire trick is built on one law, the power of a product rule:

a^{n} \cdot b^{n} = (a \cdot b)^{n}.

This is the fourth law in the list of the six exponent laws, and it is the most under-used one in exams. Every other law has obvious uses — the product law a^m \cdot a^n = a^{m+n}, the power-of-a-power (a^m)^n = a^{mn}, and so on. But a^n \cdot b^n = (ab)^n is the one that lets you combine different bases into a single base, which is exactly what you need when the exponents are already the same.

Why the law works: a^n \cdot b^n is n copies of a multiplied by n copies of b. Group them in pairs: (a \cdot b)(a \cdot b) \cdots (a \cdot b) with n pairs. That is (ab)^n. The commutativity of multiplication is what lets you interleave the as and bs.

When you spot that 2^{10} and 3^{10} both have the same exponent 10, you can fuse them into (2 \cdot 3)^{10} = 6^{10}. Now the numerator 6^{10} and the denominator 6^{10} match exactly, and they cancel to 1. Total arithmetic done: none.

The recognition pattern

Whenever you see a product or quotient of powers with matching exponents, the right move is to fuse the bases. Here are the three flavours:

Flavour 1 — fuse a product. a^n \cdot b^n = (ab)^n.

Flavour 2 — fuse a quotient. \dfrac{a^n}{b^n} = \left(\dfrac{a}{b}\right)^n.

Flavour 3 — fuse then cancel. If a \cdot b = c, then \dfrac{a^n \cdot b^n}{c^n} = \dfrac{(ab)^n}{c^n} = 1.

The original problem is Flavour 3 with a = 2, b = 3, c = 6.

The interactive recognition

Interactive fuse-and-cancel with a draggable exponent n A draggable point N on a horizontal axis from one to twelve. Live readouts show two to the n, three to the n, the product of those, six to the n, and the quotient of the two — which always equals one. As the reader drags n, the first two numbers grow dramatically but the ratio stays locked at one, demonstrating the shortcut. 1 6 12 ← drag n to change the exponent →
Slide $n$ from $1$ to $12$. The numbers $2^n$, $3^n$, and $6^n$ all grow explosively — by $n = 12$, $6^{12}$ is more than two billion. But the ratio $(2^n \cdot 3^n)/6^n$ stays exactly $1$ for every $n$. The shortcut is not approximate; it is an identity.

A wider family of problems

Once you see the move, it works on any problem of this shape. A few examples you might actually meet in a JEE paper:

Example 1. Simplify \dfrac{4^{7} \cdot 25^{7}}{100^{7}}.

Because 4 \cdot 25 = 100, the numerator is (4 \cdot 25)^7 = 100^7, and the quotient is 1.

Example 2. Simplify \dfrac{15^{20}}{3^{20} \cdot 5^{20}}.

Because 3 \cdot 5 = 15, the denominator is (3 \cdot 5)^{20} = 15^{20}. The quotient is 1.

Example 3. Simplify 2^{10} \cdot 5^{10}.

Because 2 \cdot 5 = 10, the product is (2 \cdot 5)^{10} = 10^{10}. Simplified without computing 2^{10} or 5^{10} separately.

Example 4. Simplify \dfrac{8^{15}}{2^{15} \cdot 4^{15}}.

Because 2 \cdot 4 = 8, the denominator is 8^{15}. The quotient is 1.

Example 5. Evaluate \left(\dfrac{9}{16}\right)^{5} \cdot \left(\dfrac{16}{9}\right)^{5}.

Because the exponents match, fuse: \left(\dfrac{9}{16} \cdot \dfrac{16}{9}\right)^{5} = 1^{5} = 1.

In every one of these, brute-force expansion would be a disaster. The shortcut makes each of them a one-line answer.

Worked example — a slightly harder case

Problem. Simplify \dfrac{18^{10}}{2^{10} \cdot 3^{20}}.

Watch the exponents. 18^{10} has exponent 10. 2^{10} has exponent 10. 3^{20} has exponent 20 — different from the others. The fuse trick needs matching exponents, so you must first peel the 3^{20} apart.

3^{20} = (3^{2})^{10} = 9^{10}.

Why this works: power-of-a-power says (a^m)^n = a^{mn}. Reading it backwards, a^{mn} = (a^m)^n. So 3^{20} = 3^{2 \cdot 10} = (3^2)^{10} = 9^{10}. This is the law that lets you split or merge exponents as needed.

Now all three pieces have exponent 10, and we can fuse:

\frac{18^{10}}{2^{10} \cdot 9^{10}} = \frac{18^{10}}{(2 \cdot 9)^{10}} = \frac{18^{10}}{18^{10}} = 1.

Answer. 1.

The one-line takeaway

If you see a product or quotient of powers and you notice the exponents agree, regroup the bases using a^n \cdot b^n = (ab)^n or \dfrac{a^n}{b^n} = (a/b)^n. The arithmetic disappears and the simplification becomes obvious.

If the exponents do not agree, the trick is either to use the power-of-a-power law to make them match (as in the worked example), or to use the product law a^m \cdot a^n = a^{m+n} to combine powers of the same base. Between those three laws — product, quotient-of-same-base, and power-of-a-product — almost every exam exponent simplification is a short pattern-match away.

The reason this trick is worth practising until it becomes automatic is that exam time is short. A student who has to expand 2^{10} and 3^{10} by hand will lose five minutes. A student who recognises (2 \cdot 3)^{10} = 6^{10} answers in ten seconds. Over a full JEE paper, those savings add up to a letter-grade's worth of marks.

Related: Exponents and Powers · Live Tile Checker: Drag m and n and Watch the Three Exponent Laws Hold · Tile-View Proof of the Three Core Exponent Laws · Add or Multiply the Exponents? The Deep Reason the Two Laws Feel Inconsistent