Power Play (Exponents) | FIO

Question 6

Write the given number as the product of two or more powers in three different ways. The powers can be any integers.

(i) $64^3$ (ii) $192^8$ (iii) $32^{-5}$

Solution

Understand the Question

A number with an exponent can be rewritten in multiple forms using fundamental laws of exponents:
- Power of a power: $(a^m)^n = a^{m \times n}$
- Power of a product: $(a \times b)^n = a^n \times b^n$
- Product of powers with the same base: $a^m \times a^n = a^{m+n}$
By expressing the base as different products or powers of prime/composite factors, we can represent the expression as a product of powers in multiple ways.

(i) Write $64^3$ as the product of two or more powers in three different ways.

Step 1 · Express $64^3$ in Three Different Forms

Prime factorisation of the base $64$ :

\begin{aligned} 64 &= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\ &= 2^6 \end{aligned}

First way (Base 2): Using $(a^m)^n = a^{m \times n}$ :

\begin{aligned} 64^3 &= (2^6)^3 \\ &= 2^{6 \times 3} \\ &= 2^{18} \end{aligned}

Second way (Base 8): Since $64 = 8^2$ :

\begin{aligned} 64^3 &= (8^2)^3 \\ &= 8^{2 \times 3} \\ &= 8^6 \end{aligned}

Third way (Base 4): Since $64 = 4^3$ :

\begin{aligned} 64^3 &= (4^3)^3 \\ &= 4^{3 \times 3} \\ &= 4^9 \end{aligned}

Answer

(i) $2^{18}$ , $8^6$ , and $4^9$

(ii) Write $192^8$ as the product of two or more powers in three different ways.

Step 1 · Express $192^8$ in Three Different Forms

Prime factorisation of $192$ :

\begin{aligned} 192 &= 2 \times 96 \\ &= 2 \times 2 \times 48 \\ &= 2 \times 2 \times 2 \times 24 \\ &= 2 \times 2 \times 2 \times 2 \times 12 \\ &= 2 \times 2 \times 2 \times 2 \times 2 \times 6 \\ &= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \\ &= 2^6 \times 3^1 \end{aligned}

First way: Using $(a \times b)^n = a^n \times b^n$ :

\begin{aligned} 192^8 &= (2^6 \times 3^1)^8 \\ &= (2^6)^8 \times (3^1)^8 \\ &= 2^{6 \times 8} \times 3^{1 \times 8} \\ &= 2^{48} \times 3^8 \end{aligned}

Second way: Using $192 = 64 \times 3$ :

\begin{aligned} 192^8 &= (64 \times 3)^8 \\ &= 64^8 \times 3^8 \end{aligned}

Third way: Writing $2^6 = 2^3 \times 2^3$ , so $192 = 2^3 \times 2^3 \times 3$ : $192^8 = (2^3 \times 2^3 \times 3)^8$

Answer

(ii) $2^{48} \times 3^8$ , $64^8 \times 3^8$ , and $(2^3 \times 2^3 \times 3)^8$

(iii) Write $32^{-5}$ as the product of two or more powers in three different ways.

Step 1 · Express $32^{-5}$ in Three Different Forms

Prime factorisation of $32$ :

\begin{aligned} 32 &= 2 \times 2 \times 2 \times 2 \times 2 \\ &= 2^5 \end{aligned}

First way: Using $(a^m)^n = a^{m \times n}$ :

\begin{aligned} 32^{-5} &= (2^5)^{-5} \\ &= 2^{5 \times (-5)} \\ &= 2^{-25} \end{aligned}

Second way: Writing $32 = 2^5 = 2^3 \times 2^2$ : $32^{-5} = (2^3 \times 2^2)^{-5}$

Third way: Using different bases where $32 = 4^3 \times 2^{-1}$ :

\begin{aligned} 4^3 \times 2^{-1} &= (2^2)^3 \times 2^{-1} \\ &= 2^{2 \times 3} \times 2^{-1} \\ &= 2^6 \times 2^{-1} \\ &= 2^{6 + (-1)} \\ &= 2^{6-1} \\ &= 2^5 \end{aligned}

Therefore: $32^{-5} = (4^3 \times 2^{-1})^{-5}$

Answer

(iii) $2^{-25}$ , $(2^3 \times 2^2)^{-5}$ , and $(4^3 \times 2^{-1})^{-5}$

Common Mistakes

Multiplying vs. Adding Exponents: Confusing $(a^m)^n = a^{m \times n}$ with $a^m \times a^n = a^{m+n}$ .
Distributing Exponents Incorrectly: Forgetting to apply the outer power to all factors inside parentheses, e.g. $(a \times b)^n = a^n \times b^n$ .
Sign Errors with Negative Exponents: Making calculation errors when multiplying a positive exponent by a negative exponent, e.g., $5 \times (-5) = -25$ , not $+25$ .

More questions in FIO