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243 LearnersLast updated on December 10, 2025

Picture stacks of blocks. Multiplying exponents is just combining piles! When the block types match, don't overcomplicate it—simply combine the totals. If you take a pile of three and stack it on a pile of four, what's the new height? You just added exponents!

Multiplication of exponents, commonly known as the Product Rule, is a mathematical operation used to combine terms that share the same base. When you multiply two exponential expressions with identical bases, you retain that base and add the exponents together to determine the final power. This simplifies complex algebraic and numerical expressions into a single term.
Examples:
Exponents are a concise mathematical notation for expressing repeated multiplication, where a "base" number is multiplied by itself a specific number of times, as indicated by the superscript. This notation simplifies the representation of large values and patterns without requiring long factor strings. Clear instances of this concept include \(2^5 = 32, 10^3 = 1000, 4^3 = 64\), and \(9^2 = 81\).
In multiplying exponents with the same base, we keep the base the same and only the exponents get added. Let ap and aq be the two terms, where ‘a’ is the base and the exponents are ‘p’ and ‘q.’
If the terms are given as ap and aq, we multiply them as:
\(a^p \times a^q = a^{(p+q)} \)
Example:
Find the product of \(\mathbf{3^2 \times 5^2}\)
Now substitute the values in the expression \(a^p \times b^p = (a \times b)^p\):
\(3^2 \times 5^2 = (3 \times 5)^2 = 15^2\)
Since \(15^2\) is '\(15 \times 15\)', we get the product as 225


Multiplying exponents with different bases depends on two scenarios:
Different bases with the same exponent
For terms with different bases and the same exponent, the expression will be in the form: \( a^p x b^p\). Here, ‘a’ and ‘b’ are the bases, where ‘p’ will be the exponent for both the terms. To get the product, first multiply the base and then apply the exponent. Hence, we can write \(a^p \times b^p = (a \times b)^p \).
Example:
Find the product of 22 × 42
Now substitute the values in the expression \(a^p \times b^p = (a \times b)^p \):
\(2^2 \times 4^2 = (2 \times 4)^2 = 8^2 \)
Since 82 is ‘\(8 × 8\)’, we get the product as 64
Different bases with different exponents
When the given terms differ in both base and exponent, they are solved separately. If the given terms are ap and bq, their product will be (ap) × (bq).
Example:
Find the product of \((4^2) \times (5^3) \)
First, find the values of 42 and 53. After finding the product of each, multiply them together to obtain the final product.
\(4^2 = 4 \times 4 = 16 \)
\(5^3 = 5 \times 5 \times 5 = 125 \)
Therefore,
\(4^2 \times 5^3 = 16 \times 125 \)
\(16 × 125 = 2000\)
While multiplying terms with negative exponents, We can take the reciprocal of the base. For example,\(\) 4-2 can be expressed as \(\frac{1}{4^2} \)
There are three cases to be followed while you multiply the terms with negative exponents. Let’s discuss them in more detail.
Case 1: The formula to be used when we have the same bases, but different negative exponents is \(a^{-p} \times a^{-q} = \frac{1}{a^{p+q}} \)
Example:
Find the product of \(2^{-1} \times 2^{-2} \)
According to the formula, we find the product as
\(2^{-1} \times 2^{-2} = 2^{-(1+2)} = 2^{-3} = \frac{1}{8} = 0.125 \)
Case 2: When the bases are different, but the negative exponents are the same,
use the formula \(a^{-p} \times b^{-p} = \frac{1}{(a \times b)^p} \)
Example:
Find the product of \(3^{-2} \) and \(4^{-2} \)
According to the formula, we find the product as
\(3^{-2} \times 4^{-2} = \frac{1}{(3 \times 4)^2} = \frac{1}{12^2} = \frac{1}{144} \approx 0.006944 \)
Case 3: When both the bases and negative exponents are different, use the formula \(a^{-p} \times b^{-q} = \frac{1}{a^p} \times \frac{1}{b^q} \)’
Example:
Multiply \(2^{-2}\) and \(3^{-4}\)
According to the formula, we find the product as
\(2^{-2} \times 3^{-4} = \frac{1}{2^2} \times \frac{1}{3^4} = \frac{1}{4} \times \frac{1}{81} = \frac{1}{4 \times 81} = \frac{1}{324} \approx 0.00309 \)
The same exponent rules apply when we have to multiply exponents with variables.
While multiplying the exponents with a square root, the base remains a square root and the same exponential rules apply. The ‘√’ symbol is used to express the square root of a number. Hence, √b can be written in its radical form as b1/2.
To write the given exponential expression into radical exponent, multiply the exponent with ½. For example, (√b)n is expressed as (b1/2)n which gives bn/2.
Given below are the rules to be followed while multiplying the exponents with square roots.
Rule 1: Add the exponents when both the square root bases are the same
Example:
Find the product of (√3)2 and (√3)3
\((\sqrt{3})^2 \times (\sqrt{3})^3 = (\sqrt{3})^{2+3} = (\sqrt{3})^5 \)
We know that \((\sqrt{b})^n = (b^{1/2})^n = b^{n/2} \). Therefore, (√3)5 can be expressed as (√3)5/2.
Rule 2: If the exponents are the same and the square root bases are different, the bases are multiplied first
Example:
Find the product of (√5)2 and (√2)2
\((\sqrt{5})^2 \times (\sqrt{2})^2 = (\sqrt{5} \times \sqrt{2})^2 = (\sqrt{10})^2 = 10^{2/2} = 10^1 = 10 \)
Rule 3: If the base of the square root is different from the exponent, then they are calculated separately and then multiplied together.
Example:
Find the product of (√5)3 and (√3)5
\((\sqrt{5})^3 = 5^{3/2} \)
\((\sqrt{3})^5 = 3^{5/2} \)
So the product will be \((\sqrt{5})^2 \times (\sqrt{3})^5 \)
Here, the base will be a fraction. For e.g., a/bn, where ‘a/b’ is the base and ‘n’ is the exponent.
The rules to multiply the exponents with fractions are given below:
Fractional exponents refer to a term’s exponent in a fractional form. Let’s consider the term ‘am/n’. Here, ‘m/n’ is the exponent.
Check the given table to understand the rules applied while multiplying.
| Rules | Formula |
| When the bases are the same and the exponents are different, add the exponents | \(a^{\frac{m}{n}} \times a^{\frac{x}{y}} = a^{\frac{m}{n} + \frac{x}{y}} \) For example,\(2^{2/4} \times 2^{3/9} = 2^{1/2 + 1/3} = 2^{5/6} \) |
| When the bases are different, but the fractional exponents are the same, multiply the base first and then evaluate. | \(a^{m/n} \times b^{m/n} = (a \times b)^{m/n} \) For example, \(2^{3/4} \times 3^{3/4} = 6^{3/4} \) |
| When the exponents and the bases are different, calculate them separately. | \(a^{m/n} \times b^{x/y} = \left(a^{m/n}\right) \times \left(b^{x/y}\right) \) For example, \(2^{2/3} \times 3^{3/4} \approx 1.587 \times 2.279 \approx 3.62 \) |
The concept of exponents can often feel abstract because it compresses large mathematical operations into tiny superscripts. When students first encounter these problems, the instinct is usually to multiply the numbers they see rather than apply the specific rules of the operation. To bridge the gap between rote memorization and proper understanding, here are a few tips and tricks to help reinforce the logic behind the math.
Dealing with problems involving the multiplication of exponents can be tricky and lead to mistakes. Let’s discuss them further and the solutions to avoid them.
We use exponents in our daily life. Given below are some real-life applications. Let’s discuss them further:
Multiply 6^4 and 5^4
304, which is 810000.
The given terms are 64 and 54. Here the bases are different and exponents remain unchanged. Hence, multiply the bases.
\(6^4 \times 5^4 = 30^4 = 30 \times 30 \times 30 \times 30 = 810{,}000 \)
What will be the product when 2^4 and 2^5 are multiplied?
The result is 29, which is 512
Here, we have same bases and different exponents. So, we simply add the exponents.
\(2^4 \times 2^5 = 2^{4+5} = 2^9 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512 \)
Find the product of [ (2^4 × 3^4) × (4^1 × 4^3)]
244 will be the product
\((2^4 \times 3^4) = 6^4 \)
\((4^1 \times 4^3) = 4^4 \)
Therefore, \((2^4 \times 3^4) \times (4^1 \times 4^3) = 6^4 \times 4^4 = 24^4 \)
What will be the product of (√3)^2 and (√5)^2
15
\((\sqrt{3})^2 \times (\sqrt{5})^2 = (\sqrt{3} \times \sqrt{5})^2 = (\sqrt{15})^2 = 15^{2/2} = 15 \)
Multiply 2^2/3 and 2^3/4
217/12
Since the bases are the same and the exponents are different, add the exponents to get the product.
\(2^{2/3} \times 2^{3/4} = 2^{\frac{2}{3} + \frac{3}{4}} = 2^{\frac{8}{12} + \frac{9}{12}} = 2^{\frac{17}{12}} \)
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.






