Simplify (2 cubed x 3 squared x 5)/(2 x 3 to the 4th) using laws of exponents

Question

Simplify the expression using laws of exponents:

\frac{2^3 \times 3^2 \times 5}{2 \times 3^4}

Solution — Step by Step

\frac{2^3 \times 3^2 \times 5}{2^1 \times 3^4}

Note that $2 = 2^1$ and $3^4 = 3^4$ in the denominator. Writing explicit exponents makes it easier to apply the division law correctly.

Group like bases together:

= \frac{2^3}{2^1} \times \frac{3^2}{3^4} \times \frac{5}{1}

This uses the commutative and associative properties of multiplication.

The key law: $\frac{a^m}{a^n} = a^{m-n}$

\frac{2^3}{2^1} = 2^{3-1} = 2^2 = 4

\frac{3^2}{3^4} = 3^{2-4} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}

And $5$ stays as $5$ (no matching base in denominator).

= 4 \times \frac{1}{9} \times 5 = \frac{4 \times 5}{9} = \frac{20}{9}

The simplified form is $\mathbf{\dfrac{20}{9}}$ .

Why This Works

The division law $a^m \div a^n = a^{m-n}$ comes from expanding: $\frac{a \cdot a \cdot a \cdots (m\text{ times})}{a \cdot a \cdots (n\text{ times})} = a^{m-n}$ after cancellation. When $m < n$ , you get a negative exponent, which means a fraction: $a^{-k} = \frac{1}{a^k}$ .

The critical rule: you can only apply the division law to the same base. You cannot combine $2^3$ and $3^4$ using this law — they must be handled separately.

Alternative Method

Expand fully and cancel:

\frac{2 \times 2 \times 2 \times 3 \times 3 \times 5}{2 \times 3 \times 3 \times 3 \times 3}

Cancel one 2 and two 3s from numerator with the denominator:

= \frac{2 \times 2 \times 5}{3 \times 3} = \frac{20}{9}

Same answer. For small exponents, this “expand and cancel” method is quick and builds intuition. For large exponents (like $2^{15}/2^{12}$ ), use the law — don’t expand.

When the exponent in the numerator is smaller than in the denominator (like $3^2/3^4$ ), the result is a negative exponent: $3^{2-4} = 3^{-2}$ . Always convert negative exponents to fractions at the end: $3^{-2} = 1/9$ . Leaving negative exponents in a “simplified” answer is considered incomplete in CBSE marking.

Common Mistake

Students often subtract exponents across different bases: they write $\frac{2^3 \times 3^2}{2 \times 3^4} = \frac{6^5}{6^5} = 1$ by treating $2^3 \times 3^2$ as $6^5$ . This is completely wrong. $2^3 \times 3^2 = 8 \times 9 = 72$ , not $6^5 = 7776$ . The laws of exponents apply within the same base only. $a^m \times b^m = (ab)^m$ (same exponent, different bases), but $a^m \times a^n = a^{m+n}$ (same base, different exponents). Never mix these up.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next