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