? Writing Numbers and Expressions as Single Powers Worksheet - Free Printable

Here are the solutions to the problems shown on the right-hand page (Page B).

1. Write each number as a single power of the number in brackets.

a. $16$ $(2)$
* Think: How many times do we multiply 2 by itself to get 16?
* $2 \times 2 = 4$
* $4 \times 2 = 8$
* $8 \times 2 = 16$
* That is 4 twos.
* Answer: $2^4$

b. $64$ $(2)$
* We know from part (a) that $2^4 = 16$. Let's keep multiplying by 2.
* $16 \times 2 = 32$ ($2^5$)
* $32 \times 2 = 64$ ($2^6$)
* Answer: $2^6$

c. $27$ $(3)$
* Think: How many times do we multiply 3 by itself to get 27?
* $3 \times 3 = 9$
* $9 \times 3 = 27$
* That is 3 threes.
* Answer: $3^3$

d. $625$ $(5)$
* Think: How many times do we multiply 5 by itself to get 625?
* $5 \times 5 = 25$
* $25 \times 5 = 125$
* $125 \times 5 = 625$
* That is 4 fives.
* Answer: $5^4$

e. $\frac{1}{9}$ $(3)$
* First, look at the whole number 9. We know $3 \times 3 = 9$, so $9 = 3^2$.
* Because it is a fraction $\frac{1}{9}$, the power must be negative.
* Rule: $\frac{1}{x^n} = x^{-n}$
* Answer: $3^{-2}$

f. $10\,000$ $(10)$
* Count the zeros. There are 4 zeros.
* This means we multiply 10 by itself 4 times.
* Answer: $10^4$

g. $\frac{1}{81}$ $(3)$
* First, look at 81. From question 1(d), we saw that $3 \times 3 \times 3 \times 3 = 81$. So, $81 = 3^4$.
* Because it is $\frac{1}{81}$, we make the power negative.
* Answer: $3^{-4}$

h. $\sqrt{2}$ $(2)$
* A square root is the same as the power of one-half.
* Rule: $\sqrt{x} = x^{\frac{1}{2}}$
* Answer: $2^{\frac{1}{2}}$

i. $\sqrt[3]{5}$ $(5)$
* A cube root (the little 3 outside the root) is the same as the power of one-third.
* Rule: $\sqrt[n]{x} = x^{\frac{1}{n}}$
* Answer: $5^{\frac{1}{3}}$

---

2. Use index laws to write each number as a single power of 2.

a. $16^4$
* First, change 16 into a power of 2. We know $16 = 2^4$.
* So, replace 16 with $2^4$: $(2^4)^4$.
* Rule: When you have a power inside a bracket and a power outside, you multiply them ($4 \times 4$).
* Answer: $2^{16}$

b. $8 \times 4^3$
* Change everything to base 2.
* $8 = 2^3$
* $4 = 2^2$, so $4^3 = (2^2)^3 = 2^6$ (multiply powers $2 \times 3$).
* Now the problem is: $2^3 \times 2^6$.
* Rule: When multiplying with the same base, you add the powers ($3 + 6$).
* Answer: $2^9$

c. $4^2 \times 32^3$
* Change everything to base 2.
* $4 = 2^2$, so $4^2 = (2^2)^2 = 2^4$ (multiply powers $2 \times 2$).
* $32 = 2^5$, so $32^3 = (2^5)^3 = 2^{15}$ (multiply powers $5 \times 3$).
* Now the problem is: $2^4 \times 2^{15}$.
* Add the powers: $4 + 15 = 19$.
* Answer: $2^{19}$

d. $\frac{1}{2} \times 64^2$
* Change everything to base 2.
* $\frac{1}{2} = 2^{-1}$
* $64 = 2^6$, so $64^2 = (2^6)^2 = 2^{12}$ (multiply powers $6 \times 2$).
* Now the problem is: $2^{-1} \times 2^{12}$.
* Add the powers: $-1 + 12 = 11$.
* Answer: $2^{11}$

e. $(\sqrt{2})^5$
* Change the square root to a fraction power: $\sqrt{2} = 2^{\frac{1}{2}}$.
* So, $(2^{\frac{1}{2}})^5$.
* Multiply the powers: $\frac{1}{2} \times 5 = \frac{5}{2}$ (or $2.5$).
* Answer: $2^{\frac{5}{2}}$ or $2^{2.5}$

f. $\frac{1}{2} \times 16^2$
* Change everything to base 2.
* $\frac{1}{2} = 2^{-1}$
* $16 = 2^4$, so $16^2 = (2^4)^2 = 2^8$ (multiply powers $4 \times 2$).
* Now the problem is: $2^{-1} \times 2^8$.
* Add the powers: $-1 + 8 = 7$.
* Answer: $2^7$

---

3. Given that $2^{2y+3} \times 3^y$, express $y$ in terms of $x$.

*(Note: The image text for Question 3 appears to contain a typo or missing information, as the variable '$x$' is not present in the expression provided. Usually, these questions say "Given that [Expression] = $x$". Assuming the standard format where the expression equals $x$, here is the solution:)*

Assumption: The question intends to say: Given that $x = 2^{2y+3} \times 3^y$ (or potentially just involving base 2 if the $3^y$ is a typo for another power of 2, but let's solve for the most likely algebraic intent).

However, looking closely at typical worksheets of this level, Question 3 often relates to the previous section (Base 2). If the question was actually $2^{2y+3} = x$, here is how you solve it:

1. Take the log of both sides or rewrite in exponential form.
2. $\log_2(x) = 2y + 3$
3. Subtract 3 from both sides: $\log_2(x) - 3 = 2y$
4. Divide by 2: $y = \frac{\log_2(x) - 3}{2}$

Alternative Interpretation (Simpler Algebra):
If the question meant $2^{2y+3} = 2^x$ (equating powers of 2):
1. Since the bases are the same, the exponents must be equal.
2. $2y + 3 = x$
3. Subtract 3: $2y = x - 3$
4. Divide by 2: $y = \frac{x - 3}{2}$

*(Without the explicit "= x" or similar constraint visible in the crop, the most standard algebraic manipulation for "express y in terms of x" given a power equation usually implies equating exponents or using logs. Based on the difficulty of Q1 and Q2, the linear exponent solution is most likely intended if there was an equality to a power of 2).*

Final Answer for Q3 (assuming $2^{2y+3} = 2^x$):
$y = \frac{x - 3}{2}$