Question 6

We want to write the given number as a product of two or more powers in three different ways. We will use the rules of exponents.

Step 1 — Expressing $64^3$ in different ways

First, let us find the prime factorization of the base number, 64. $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $= 2^6$ So, we can write $64^3$ using the base 2. The rule $(a^m)^n = a^{m \times n}$ means we multiply the exponents. $64^3 = (2^6)^3$ $= 2^{6 \times 3}$

$\boxed{2^{18}}$ We can also express 64 using other bases that are powers of 2. For example, $64 = 8^2$, because $8 = 2^3$. $64^3 = (8^2)^3$ $= 8^{2 \times 3}$ $\boxed{8^6}$ Another way to express 64 is using base 4. We know $64 = 4^3$, because $4 = 2^2$. $64^3 = (4^3)^3$ $= 4^{3 \times 3}$ $\boxed{4^9}$

Step 2 — Expressing $192^8$ in different ways

Let us find the prime factorization of the base number, 192. $192 = 2 \times 96$ $= 2 \times 2 \times 48$ $= 2 \times 2 \times 2 \times 24$ $= 2 \times 2 \times 2 \times 2 \times 12$ $= 2 \times 2 \times 2 \times 2 \times 2 \times 6$ $= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$ $= 2^6 \times 3^1$ Now, we can apply the outer exponent, 8, to this product. The rule $(a \times b)^n = a^n \times b^n$ means we apply the exponent to each factor. $192^8 = (2^6 \times 3^1)^8$ $= (2^6)^8 \times (3^1)^8$ $= 2^{6 \times 8} \times 3^{1 \times 8}$

$\boxed{2^{48} \times 3^8}$ For a second way, we can express 192 as a product of a composite number and a prime number. We know $192 = 64 \times 3$. $192^8 = (64 \times 3)^8$ $= 64^8 \times 3^8$ $\boxed{64^8 \times 3^8}$ For a third way, we can split the prime factors of 192 differently. We know $192 = 2^6 \times 3$. We can write $2^6$ as $2^3 \times 2^3$. So, we can write $192$ as $2^3 \times 2^3 \times 3$. $192^8 = (2^3 \times 2^3 \times 3)^8$ $\boxed{(2^3 \times 2^3 \times 3)^8}$

Step 3 — Expressing $32^{-5}$ in different ways

First, let us find the prime factorization of the base number, 32. $32 = 2 \times 2 \times 2 \times 2 \times 2$ $= 2^5$ Now, we can write $32^{-5}$ using the base 2. The rule $(a^m)^n = a^{m \times n}$ means we multiply the exponents. $32^{-5} = (2^5)^{-5}$ $= 2^{5 \times (-5)}$

$\boxed{2^{-25}}$ For a second way, we can express the base 32 as a product of powers of 2. We know $32 = 2^5$. We can write $2^5$ as $2^3 \times 2^2$. So, we can write $32$ as $2^3 \times 2^2$. $32^{-5} = (2^3 \times 2^2)^{-5}$ $\boxed{(2^3 \times 2^2)^{-5}}$ For a third way, we can express the base 32 as a product of powers using different bases. We need to find two powers whose product is 32. Let us try $4^3 \times 2^{-1}$. We check this: $4^3 \times 2^{-1} = (2^2)^3 \times 2^{-1} = 2^{2 \times 3} \times 2^{-1} = 2^6 \times 2^{-1}$. The rule $a^m \times a^n = a^{m+n}$ means we add the exponents when the bases are the same. $2^6 \times 2^{-1} = 2^{6 + (-1)}$ $= 2^{6-1}$ $= 2^5$ So, $4^3 \times 2^{-1}$ is indeed equal to 32. Now we can write $32^{-5}$ using this expression for the base. $32^{-5} = (4^3 \times 2^{-1})^{-5}$ $\boxed{(4^3 \times 2^{-1})^{-5}}$

Answer

(i) $64^3$ can be written as: $2^{18}$ $8^6$ $4^9$ (ii) $192^8$ can be written as: $2^{48} \times 3^8$ $64^8 \times 3^8$ $(2^3 \times 2^3 \times 3)^8$ (iii) $32^{-5}$ can be written as: $2^{-25}$ $(2^3 \times 2^2)^{-5}$ $(4^3 \times 2^{-1})^{-5}$