map 4c1
unit 6 quiz: exponential expressions
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total: 2/20
instructions: please provide full, polished solutions for each in the spaces
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1. write each of the following as a single power. then evaluate.
(a) $3^3 \times 3^4$
(b) $\frac{5^{10}}{5^7}$ (2 mks.)
(c) $\frac{(-8)^3 \times (-8)^5}{(-8)^9}$
(d) $\frac{(2^2)^8 \times 2^2}{(2^3)^4}$ (4 mks.)
2. simplify each to a power that has positive exponents only. then evaluate.
(a) $7^{-4}$
(b) $(-12)^{-3}$ (4 mks.)
(over..)
unit 6 quiz

Question

Accepted Answer

### Problem 1(a): $ 3^3 \times 3^4 $
# Explanation:
## Step 1: Use exponent rule $ a^m \times a^n = a^{m + n} $
For $ 3^3 \times 3^4 $, we add the exponents since the base is the same. So $ 3^{3 + 4} = 3^7 $.
## Step 2: Evaluate $ 3^7 $
$ 3^7 = 3\times3\times3\times3\times3\times3\times3 = 2187 $.
# Answer: Single power is $ 3^7 $, value is $ 2187 $

### Problem 1(b): $ \frac{5^{10}}{5^7} $
# Explanation:
## Step 1: Use exponent rule $ \frac{a^m}{a^n} = a^{m - n} $
For $ \frac{5^{10}}{5^7} $, subtract the exponents. So $ 5^{10 - 7} = 5^3 $.
## Step 2: Evaluate $ 5^3 $
$ 5^3 = 5\times5\times5 = 125 $.
# Answer: Single power is $ 5^3 $, value is $ 125 $

### Problem 1(c): $ \frac{(-8)^3 \times (-8)^5}{(-8)^9} $
# Explanation:
## Step 1: Use $ a^m \times a^n = a^{m + n} $ for numerator
$ (-8)^3 \times (-8)^5 = (-8)^{3 + 5} = (-8)^8 $.
## Step 2: Use $ \frac{a^m}{a^n} = a^{m - n} $
Now, $ \frac{(-8)^8}{(-8)^9} = (-8)^{8 - 9} = (-8)^{-1} $. But we can also do it in one step: $ (-8)^{3 + 5 - 9}=(-8)^{-1} $. However, to write as a single power (and then evaluate, but let's first simplify exponents). Wait, actually, the rule for same base multiplication and division: $ \frac{(-8)^3 \times (-8)^5}{(-8)^9}=(-8)^{3 + 5 - 9}=(-8)^{-1} $. But if we want positive exponent, $ (-8)^{-1}=\frac{1}{-8}=-\frac{1}{8} $. But the question says "write as a single power then evaluate". So first, single power: $ (-8)^{-1} $ or we can also think of it as $ \frac{1}{(-8)^1} $. But let's follow the exponent rules.
Wait, maybe I made a mistake earlier. Let's re - do:
$ (-8)^3\times(-8)^5 = (-8)^{3 + 5}=(-8)^8 $ (since multiplying two negative numbers gives positive, and exponent addition). Then $ \frac{(-8)^8}{(-8)^9}=(-8)^{8 - 9}=(-8)^{-1} $. Now, evaluating $ (-8)^{-1}=\frac{1}{-8}=-\frac{1}{8} $. But if we consider the single power first: the single power is $ (-8)^{-1} $, and its value is $ -\frac{1}{8} $.
# Answer: Single power is $ (-8)^{-1} $ (or $ \frac{1}{-8} $), value is $ -\frac{1}{8} $

### Problem 1(d): $ \frac{(2^2)^8 \times 2^2}{(2^3)^4} $
# Explanation:
## Step 1: Use $ (a^m)^n = a^{m\times n} $
For $ (2^2)^8 = 2^{2\times8}=2^{16} $, and $ (2^3)^4 = 2^{3\times4}=2^{12} $.
## Step 2: Use $ a^m \times a^n = a^{m + n} $ for numerator
$ 2^{16} \times 2^2 = 2^{16 + 2}=2^{18} $.
## Step 3: Use $ \frac{a^m}{a^n} = a^{m - n} $
$ \frac{2^{18}}{2^{12}} = 2^{18 - 12}=2^6 $.
## Step 4: Evaluate $ 2^6 $
$ 2^6 = 64 $.
# Answer: Single power is $ 2^6 $, value is $ 64 $

### Problem 2(a): $ 7^{-4} $
# Explanation:
## Step 1: Use $ a^{-n}=\frac{1}{a^n} $
$ 7^{-4}=\frac{1}{7^4} $.
## Step 2: Evaluate $ 7^4 $
$ 7^4 = 7\times7\times7\times7 = 2401 $. So $ 7^{-4}=\frac{1}{2401} $.
# Answer: Simplified to $ \frac{1}{7^4} $, value is $ \frac{1}{2401} $

### Problem 2(b): $ (-12)^{-3} $
# Explanation:
## Step 1: Use $ a^{-n}=\frac{1}{a^n} $
$ (-12)^{-3}=\frac{1}{(-12)^3} $.
## Step 2: Evaluate $ (-12)^3 $
$ (-12)^3=(-12)\times(-12)\times(-12)= - 1728 $. So $ (-12)^{-3}=\frac{1}{-1728}=-\frac{1}{1728} $.
# Answer: Simplified to $ \frac{1}{(-12)^3} $, value is $ -\frac{1}{1728} $

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QUESTION IMAGE

Question

Explanation:

Problem 1(a): \( 3^3 \times 3^4 \)

Step 1: Use exponent rule \( a^m \times a^n = a^{m + n} \)

Step 2: Evaluate \( 3^7 \)

Step 1: Use exponent rule \( \frac{a^m}{a^n} = a^{m - n} \)

Step 2: Evaluate \( 5^3 \)

Step 1: Use \( a^m \times a^n = a^{m + n} \) for numerator

Step 2: Use \( \frac{a^m}{a^n} = a^{m - n} \)

Answer:

Problem 1(b): \( \frac{5^{10}}{5^7} \)