Question
Sumnnative aSSeSSMnent: EX Sumplify the following, without using a calculat 1. \( \frac{4^{a}-2^{2 a+1}}{\left(2^{a}\right)^{2}-2^{a+3} \cdot 2^{a}} \) 2. \( \left(\frac{1}{3^{n-1}} \frac{1}{\left.3^{n+1}\right)^{\frac{1}{n}}}\right. \) 3. \( \frac{\sqrt{a}-a}{a^{3}-\sqrt{a^{5}}} \) 4. \( \frac{9^{2 n+1} \cdot 6^{2 n-3}}{3^{5 n-2} \cdot 2 \cdot 3^{n} \cdot 4^{n-2}} \) 5. \( \sqrt[3]{27^{2}}-\frac{2}{8^{-\frac{2}{3}}}+\frac{\sqrt[5]{2}}{4^{-\frac{2}{5}}} \) 6. \( \frac{3^{2 x+1}+9^{x}}{3^{x} \cdot 3^{x+1}-\left(3^{x}\right)^{2}} \) 7. \( \frac{3 x^{2}\left(3^{x-1}\right)^{x+1}}{9^{x^{x+1}}} \) 8. \( \frac{2^{x+1}-3 \cdot 2^{x-1}}{2^{x}-2^{x-2}} \) \( 3^{x}-\left(\frac{1}{2}\right)^{x} \)
Ask by Crawford Cook. in South Africa
Jan 22,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
1. \( \frac{4^{a}-2^{2 a+1}}{\left(2^{a}\right)^{2}-2^{a+3} \cdot 2^{a}} = \frac{1}{7} \)
2. \( \left(\frac{1}{3^{n-1}} \cdot \frac{1}{(3^{n+1})^{\frac{1}{n}}}\right) = \frac{1}{3^{n + \frac{1}{n}}} \)
3. \( \frac{\sqrt{a}-a}{a^{3}-\sqrt{a^{5}}} = \frac{1 - a^{1/2}}{a^2(a^{3/2} - 1)} \)
4. \( \frac{9^{2 n+1} \cdot 6^{2 n-3}}{3^{5 n-2} \cdot 2 \cdot 3^{n} \cdot 4^{n-2}} = 3 \)
5. \( \sqrt[3]{27^{2}}-\frac{2}{8^{-\frac{2}{3}}}+\frac{\sqrt[5]{2}}{4^{-\frac{2}{5}}} = 9 - 8 + 2 = 3 \)
6. \( \frac{3^{2 x+1}+9^{x}}{3^{x} \cdot 3^{x+1}-\left(3^{x}\right)^{2}} = \frac{3^{2x+1} + 3^{2x}}{3^{2x+1} - 3^{2x}} = \frac{3^{2x}(3 + 1)}{3^{2x}(3 - 1)} = \frac{4}{2} = 2 \)
7. \( \frac{3 x^{2}\left(3^{x-1}\right)^{x+1}}{9^{x^{x+1}}} = \frac{3 x^{2} \cdot 3^{(x-1)(x+1)}}{3^{2x^{x+1}}} = \frac{3 x^{2} \cdot 3^{x^2 - 1}}{3^{2x^{x+1}}} = 3^{x^2 - 1 - 2x^{x+1}} \cdot x^{2} \)
8. \( \frac{2^{x+1}-3 \cdot 2^{x-1}}{2^{x}-2^{x-2}} = \frac{2^{x} \cdot 2 - 3 \cdot 2^{x-1}}{2^{x} - 2^{x-2}} = \frac{2^{x}(2 - \frac{3}{2})}{2^{x}(1 - \frac{1}{4})} = \frac{2^{x} \cdot \frac{1}{2}}{2^{x} \cdot \frac{3}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3} \)
9. \( 3^{x} - \left(\frac{1}{2}\right)^{x} = 3^{x} - 2^{-x} \)
Solution
Let's simplify each expression step by step.
### 1. Simplify \( \frac{4^{a}-2^{2 a+1}}{\left(2^{a}\right)^{2}-2^{a+3} \cdot 2^{a}} \)
**Step 1:** Rewrite \( 4^a \) as \( (2^2)^a = 2^{2a} \).
**Step 2:** Rewrite \( 2^{2a+1} \) as \( 2^{2a} \cdot 2^1 = 2^{2a} \cdot 2 \).
**Step 3:** The numerator becomes:
\[
2^{2a} - 2^{2a+1} = 2^{2a} - 2 \cdot 2^{2a} = 2^{2a}(1 - 2) = -2^{2a}
\]
**Step 4:** The denominator simplifies as follows:
\[
(2^a)^2 - 2^{a+3} \cdot 2^a = 2^{2a} - 2^{a+3+a} = 2^{2a} - 2^{2a+3} = 2^{2a}(1 - 2^3) = 2^{2a}(1 - 8) = -7 \cdot 2^{2a}
\]
**Step 5:** Combine the results:
\[
\frac{-2^{2a}}{-7 \cdot 2^{2a}} = \frac{1}{7}
\]
### 2. Simplify \( \left(\frac{1}{3^{n-1}} \cdot \frac{1}{(3^{n+1})^{\frac{1}{n}}}\right) \)
**Step 1:** Rewrite the second term:
\[
(3^{n+1})^{\frac{1}{n}} = 3^{\frac{n+1}{n}} = 3^{1 + \frac{1}{n}}
\]
**Step 2:** The expression becomes:
\[
\frac{1}{3^{n-1}} \cdot \frac{1}{3^{1 + \frac{1}{n}}} = \frac{1}{3^{n-1 + 1 + \frac{1}{n}}} = \frac{1}{3^{n + \frac{1}{n}}}
\]
### 3. Simplify \( \frac{\sqrt{a}-a}{a^{3}-\sqrt{a^{5}}} \)
**Step 1:** Rewrite \( \sqrt{a} \) as \( a^{1/2} \) and \( \sqrt{a^5} \) as \( a^{5/2} \).
**Step 2:** The numerator becomes:
\[
a^{1/2} - a = a^{1/2} - a^{2/2} = a^{1/2}(1 - a^{1/2})
\]
**Step 3:** The denominator becomes:
\[
a^3 - a^{5/2} = a^{5/2}(a^{3/2} - 1)
\]
**Step 4:** Combine the results:
\[
\frac{a^{1/2}(1 - a^{1/2})}{a^{5/2}(a^{3/2} - 1)} = \frac{1 - a^{1/2}}{a^2(a^{3/2} - 1)}
\]
### 4. Simplify \( \frac{9^{2 n+1} \cdot 6^{2 n-3}}{3^{5 n-2} \cdot 2 \cdot 3^{n} \cdot 4^{n-2}} \)
**Step 1:** Rewrite \( 9^{2n+1} \) as \( (3^2)^{2n+1} = 3^{4n+2} \) and \( 6^{2n-3} = (2 \cdot 3)^{2n-3} = 2^{2n-3} \cdot 3^{2n-3} \).
**Step 2:** The numerator becomes:
\[
3^{4n+2} \cdot 2^{2n-3} \cdot 3^{2n-3} = 2^{2n-3} \cdot 3^{4n+2 + 2n - 3} = 2^{2n-3} \cdot 3^{6n - 1}
\]
**Step 3:** The denominator becomes:
\[
3^{5n-2} \cdot 2 \cdot 3^{n} \cdot 4^{n-2} = 3^{5n-2 + n} \cdot 2 \cdot (2^2)^{n-2} = 3^{6n-2} \cdot 2^{1 + 2n - 4} = 3^{6n-2} \cdot 2^{2n - 3}
\]
**Step 4:** Combine the results:
\[
\frac{2^{2n-3} \cdot 3^{6n-1}}{3^{6n-2} \cdot 2^{2n-3}} = 3^{(6n-1) - (6n-2)} = 3^{1} = 3
\]
### 5. Simplify \( \sqrt[3]{27^{2}}-\frac{2}{8^{-\frac{2}{3}}}+\frac{\sqrt[5]{2}}{4^{-\frac{2}{5}}} \)
**Step 1:** Calculate \( \sqrt[3]{27^2} = \sqrt[3]{(3^3)^2} = 3^2 = 9 \).
**Step 2:** Calculate \( 8^{-\frac{2}{3}} = (2^3)^{-\frac{2}{3}} = 2^{-2} = \frac{1}{4} \), so \( \frac{2}{8^{-\frac{2}{3}}} = 2 \cdot 4 = 8 \).
**Step 3:** Calculate \( 4^{-\frac{2}{5}} = (2^2)^{-\frac{2}{5}} = 2^{-\frac{4}{5}} \), so \( \frac{\sqrt[5]{2}}{4^{-\frac{2}{5}}} = \frac{2^{\frac{1}{5}}}{2^{-\frac{4}{5}}} = 2^{\frac{1}{5} + \frac{4}{5}} = 2^{1} = 2 \).
**Step 4
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Simplify this solution Mind Expander
To simplify these expressions, it helps to remember that exponents can be manipulated using properties such as \( a^m \cdot a^n = a^{m+n} \) and \( \frac{a^m}{a^n} = a^{m-n} \). Additionally, knowing that \( 9 = 3^2 \) and \( 27 = 3^3 \) can significantly ease the process, allowing you to express everything in terms of a single base where possible. When tackling these types of problems, it’s easy to overlook elementary simplifications. Always check if you can factor common terms first or if there are opportunities to combine like bases. For example, simplifying \( 2^{a-1} \) to \( \frac{1}{2} \times 2^{a} \) may help reveal a hidden factor, or breaking an equation into smaller pieces can sometimes unveil an easier path to simplification. Happy simplifying!
