Laws of exponents

$2^3\times2^4$ looks like it needs six multiplications written out in full, but it does not. The laws of exponents are a small set of shortcut rules — add exponents here, subtract there, multiply there — that let you combine powers of the same base instantly, without ever expanding a single term. They are the computational engine of Class 8's "Power Play" chapter: every simplification problem in the chapter, however long it looks, reduces to applying these seven rules in the right order. This article compresses expanding, multiplying, dividing, powers-of-powers, the zero exponent and negative exponents into one connected picture, works two examples end to end, and flags the mix-ups that cost the most marks — mainly confusing "add the exponents" with "multiply the exponents."

The core insight

Powers of the same base combine by adding or subtracting exponents when multiplying or dividing, and by multiplying exponents when raising a power to a power — with $a^0=1$ and $a^{-n}=1/a^n$ completing the family.

The core idea

Same base, combine the exponents. For a non-zero base $a$ and integers $m,n$ , multiplying same-base powers adds exponents and dividing subtracts them:

a^m\times a^n = a^{m+n}, \qquad a^m\div a^n = a^{m-n}

where: $a$ is the common base ( $a\neq0$ ), and $m,n$ are integer exponents, positive, negative, or zero. Both rules fall straight out of what an exponent counts. $a^3\times a^2=(a\cdot a\cdot a)(a\cdot a)=a^5$ , because the exponent is simply a count of how many $a$ 's are multiplied together, and $3+2=5$ .

A power raised to a power multiplies exponents. $(a^m)^n$ means $a^m$ used as a factor $n$ times, which by the addition rule above gives $a^m$ added to itself $n$ times in the exponent:

(a^m)^n = a^{mn}

This is not a new idea — it is the multiplication law applied repeatedly, which is why remembering it alongside the addition law (rather than as a separate fact) prevents the classic mix-up of adding when you should multiply. A quick numeric check makes the difference concrete: $(2^3)^2$ means "square the number $2^3=8$ ," giving $8^2=64$ , and the shortcut agrees — $2^{3\times2}=2^6=64$ . Compare that with $2^3\times2^2$ , which instead gives $2^{3+2}=2^5=32$ . Same two numbers in the exponents, two completely different operations, two different answers — which is exactly why examiners test this pair so often.

A power distributes over a product or quotient. Raising a product or a fraction to a power hands the power to every factor:

(ab)^n = a^n b^n, \qquad \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}

where: $a,b$ are non-zero bases and $n$ is an integer. Read the product rule backwards and it becomes a second shortcut: two different bases sharing the same exponent combine into one base, e.g. $2^3\times5^3=(2\times5)^3=10^3$ . So there are two directions to watch for — same base combines by adjusting the exponent, same exponent combines by adjusting the base — and mixing them up is the single biggest source of wrong simplifications.

Zero and negative exponents complete the picture.

a^0 = 1\ (a\neq0), \qquad a^{-n} = \frac{1}{a^n}

$a^0=1$ falls out of the division law itself: $a^3\div a^3$ must equal $1$ (anything divided by itself), and the rule gives $a^{3-3}=a^0$ , so $a^0=1$ . A negative exponent is a reciprocal instruction, not a sign flip — $2^{-3}$ means $\dfrac{1}{2^3}=\dfrac{1}{8}$ , a small positive fraction, never $-8$ . The same division law explains where the negative exponent comes from in the first place: $a^2\div a^5=a^{2-5}=a^{-3}$ , but written out directly, $a^2\div a^5=\dfrac{a\times a}{a\times a\times a\times a\times a}=\dfrac{1}{a^3}$ after cancelling two factors from top and bottom — so $a^{-3}$ and $\dfrac{1}{a^3}$ are simply two names for the same quantity, not two different rules to memorise.

Figure

A flow diagram showing the seven laws of exponents grouped as combine multiply divide same base then power of a power then distribute over product or quotient then zero and negative exponent special cases

Derivation AWhy powers of a power multiply exponents

The goal is to prove $(a^m)^n=a^{mn}$ from nothing more than the meaning of an exponent and the already-established multiplication law $a^p\times a^q=a^{p+q}$ , so that the "multiply the exponents" rule is seen as a consequence rather than an extra fact to memorise separately from "add the exponents."

Step 1. Unpack what $(a^m)^n$ means. The outer exponent $n$ says "take the quantity $a^m$ and multiply it by itself $n$ times":

(a^m)^n = \underbrace{a^m\times a^m\times\cdots\times a^m}_{n\text{ factors of }a^m}

Step 2. Apply the multiplication law repeatedly. Each time two $a^m$ factors combine, their exponents add: $a^m\times a^m=a^{m+m}=a^{2m}$ , and bringing in a third factor gives $a^{2m}\times a^m=a^{2m+m}=a^{3m}$ . Continuing this $n$ times in total adds $m$ to itself $n$ times.

Step 3. Collect the result. Adding $m$ to itself $n$ times gives $m+m+\cdots+m=mn$ , so $(a^m)^n=a^{mn}$ — exactly the power-of-a-power law, now shown to be the multiplication law used $n-1$ times rather than an independent rule.

Step 4. Check it on numbers. $(3^2)^3$ should equal $3^{2\times3}=3^6=729$ . Directly: $(3^2)^3=9^3=9\times9\times9=729$ . The two routes agree, confirming the law.

Worked example

Simplify $\dfrac{5^7\times5^3}{5^4}$ , then find its value.

Step 1. The numerator has the same base $5$ on both factors, so apply the multiplication law and add exponents: $5^7\times5^3=5^{7+3}=5^{10}$ .

Step 2. The whole expression is now $\dfrac{5^{10}}{5^4}$ , again a single base, so apply the division law and subtract exponents: $5^{10-4}=5^6$ .

Step 3. Evaluate $5^6$ by multiplying it out: $5\times5\times5\times5\times5\times5=15625$ . So $\dfrac{5^7\times5^3}{5^4}=5^6=\mathbf{15625}$ .

Competitive stretch

Simplify $\dfrac{(3^4)^2\times2^{-3}}{2^{-5}\times3^5}$ in exponential form, then evaluate.

Step 1. Reduce the power-of-a-power first: $(3^4)^2=3^{4\times2}=3^8$ , turning the expression into $\dfrac{3^8\times2^{-3}}{2^{-5}\times3^5}$ .

Step 2. Group the same bases and apply the division law to each separately: $\dfrac{3^8}{3^5}\times\dfrac{2^{-3}}{2^{-5}}=3^{8-5}\times2^{-3-(-5)}=3^3\times2^2$ .

Step 3. Evaluate each power and multiply: $3^3=27$ , $2^2=4$ , and $27\times4=\mathbf{108}$ .

Common traps

Adding exponents across different bases — $2^3\times3^2$ is not $6^5$ . The addition and subtraction laws only fire when the base is identical on both sides; different bases must be simplified separately.
Swapping "add" and "multiply" — $a^m\times a^n$ adds exponents, but $(a^m)^n$ multiplies them. These look similar on the page and are the single most common slip.
Treating a negative exponent as a negative value — $2^{-3}=\dfrac{1}{8}$ , a small positive fraction, never $-8$ . A minus sign in the exponent flips to a reciprocal; it does not flip the sign of the value.
Forgetting the $a\neq0$ restriction on $a^0=1$ — the zero-exponent rule needs a non-zero base; $0^0$ is not defined at this level.
Missing the reverse shortcut — expanding $4^5\times25^5$ term by term instead of noticing the same exponent on both bases and combining directly as $(4\times25)^5=100^5$ , which is far faster and avoids arithmetic slips.

Ask the tutor: Still mixing up "add the exponents" and "multiply the exponents"? Ask the AI tutor for a side-by-side drill contrasting $a^m\times a^n$ with $(a^m)^n$ .

Why does a to the power zero always equal 1 and not 0?

Think about the division law rather than memorising the rule blindly. Dividing any non-zero number by itself always gives $1$ , so $a^3\div a^3=1$ . But the division law says $a^3\div a^3=a^{3-3}=a^0$ . For both statements to be true at once, $a^0$ must equal $1$ . The base $a$ cannot be $0$ here because $0\div0$ is not a valid division in the first place — that is exactly why the rule carries the restriction $a\neq0$ .

How is a negative exponent different from a negative number?

A negative sign in front of a number makes it smaller than zero, but a negative sign in the exponent is an instruction to take a reciprocal, not to change sign. $2^{-3}$ means "flip $2^3$ into a fraction": $2^{-3}=\dfrac{1}{2^3}=\dfrac{1}{8}$ , which is still a positive number, just a small one. The value only becomes negative if the base itself is negative and raised to an odd power, which is a completely separate situation from the exponent's sign.

Can I ever add exponents when the bases are different, like 2 cubed times 3 squared?

Not by adding exponents — that only works for a shared base. But there is a second, less obvious shortcut: if two different bases share the same exponent, you can multiply the bases together instead. For example $2^3\times5^3=(2\times5)^3=10^3=1000$ . So $2^3\times3^2$ genuinely cannot be compressed into one power — the base differs and the exponent differs — and you must evaluate each term separately: $8\times9=72$ .

Laws of exponents

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads