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Number Base System I | Week 16 Topics2 Quizzes
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Number Base System II | Week 23 Topics
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Number Base System III | Week 32 Topics1 Quiz
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Modular Arithmetic I | Week 42 Topics
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Modular Arithmetic II | Week 53 Topics
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Modular Arithmetic III | Week 64 Topics1 Quiz
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Indices I | Week 73 Topics1 Quiz
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Indices II | Week 81 Topic1 Quiz
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Logarithms I | Week 93 Topics
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Logarithms II | Week 104 Topics1 Quiz
Topic Content:
- Powers & Roots
- Evaluation Questions
Example 10.4.1:
Use logarithms to evaluate the following:
(i) \( \scriptsize \left (3.9562 \right)^3 \)
(ii) \( \scriptsize \sqrt [6] { 68.15} \)
(i) \( \scriptsize \left (3.9562 \right)^3 \)
Solution
1st Method
= \( \scriptsize \left (10^{0.5973} \right)^3 \rightarrow \left (a^x \right)^y = a^{xy} \)
= \( \scriptsize 10^{1.7919} \rightarrow \; using \; antilog \; \)
= 6.193
Answer = 61.93
2nd Method
| No | Log |
| \( \scriptsize \left (3.9562 \right)^3 \) | 0.5973 |
| 0.5793 × 3 | |
| = 1.7919 | |
| Antilog of 0.7919 = 6.193 | |
| 101 × 6.193 | |
| Ans = 61.93 |
(ii) \( \scriptsize \sqrt [6] { 68.15} \)
Solution
1st Method
\( \scriptsize \sqrt [6] { 68.15} = \left (68.15 \right)^{\frac{1}{6}} \)= \( \scriptsize 10^{1.8334 \: \div \: 6} \rightarrow \left ( \sqrt [x] {a} = a ^{ \frac{1}{x}} \right) \)
= \( \scriptsize \left (10^{0.3056} \right) \) Using antilog
= 2.021
Answer = 2.021
2nd Method
| No | Log |
| \( \scriptsize \sqrt [6] { 68.15} \) | |
| \( \scriptsize \left( 68.15 \right)^{\frac{1}{6}} \) | 1.8335 ÷ 6 |
| = 0.3056 | |
| Antilog of 0.3056 = 2.021 | |
| 100 × 2.021 | |
| Ans = 2.021 |
Example 10.4.2:
Use the log tables to evaluate the following:
(i) \( \frac{(18.6)^2 \: \times \: 9.76}{\sqrt[4]{8500}} \)
(ii) \( \sqrt[4]{\left[ \frac{43.12 \: \times \: 4.08}{3.401 \: \times \: 2.184}\right]^3} \)
(i) \( \frac{(18.6)^2 \: \times \: 9.76}{\sqrt[4]{8500}} \)
Solution
| No | Log | ||
| \( \scriptsize(18.6)^2 \) | 1.2695 × 2 = | 2.5390 | |
| 9.76 | 0.9894 | + | 0.9894 |
| 3.5284 | |||
| \(\scriptsize \sqrt[4]{(8500)} \) | |||
| \( \scriptsize \left(8500 \right)^{\frac{1}{4}}\) | 3.9294 ÷ 4 = | – | 0.9824 |
| 2.5460 | |||
| Antilog of 0.5460 = 3.516 | |||
| 102 × 3.516 | |||
| Ans = 351.6 |
(ii) \( \sqrt[4]{\left[ \frac{43.12 \: \times \: 4.08}{3.401 \: \times \: 2.184}\right]^3} \)
Solution
| No | Log | ||||
| 43.12 | 1.6347 | ||||
| 4.08 | + | 0.6107 | |||
| 2.2454 | → | 2.2454 | |||
| 3.401 | + | 0.5316 | |||
| 2.184 | 0.3393 | ||||
| 0.8709 | → | – | 0.8709 | ||
| \(\scriptsize \left[ 1.3745 \right]^3\) = | 1.3745 × 3 | ||||
| \( \scriptsize\sqrt[4]{4.1235}\) = | 4.1235 ÷ 4 | ||||
| = | 1.0309 | ||||
| Antilog of .0309 = 1.0737 | |||||
| 101 × 1.0737 | |||||
| Ans = 10.737 | |||||
| = 10.74 (2.dp) |
Evaluation
Use log and antilog tables to evaluate the following:
1. 899.5 × 83.2 × 7.5
2. \( \scriptsize 4.927^{3} \: \times \: 45.6 \)
3. \( \scriptsize 58.7 \: \div \: \sqrt[3]{8.15} \)
4. \( \scriptsize \sqrt[3]{0.07509} \)
5. \( \scriptsize (0.0099)^2 \)
6. \( \left( \frac{0.0096}{0.07812} \right)^{\frac{1}{2}} \)
7. \( \sqrt[3] {\frac{807.5 \: \times \: 45.5}{(8.932 \: \times \: 45.2)^2}} \)
8. \( \frac{(1.85)^3 \: \times \: 9.15}{\sqrt[4]{8500}} \)
9. \( \left ( \sqrt[3] {\frac{38.32 \: \times \: 2.964}{8.637\:\times \: 6.285}} \right)^2\)
Solve 10 – 12 without using log tables.
10. Simplify the following:
(a) \( \frac{\log5}{\log 0.2} \)
(b) \( \frac{\log12 \: – \: \log3}{\log 8} \)
11. Solve the following equations for y
(i) \( \scriptsize \log_y \frac{1}{16} = \: – \frac{1}{2} \)
(ii) \( \scriptsize \log_{10}(3x\:-\:1) \: -\: \log_{10}2 = 3 \)
12. Assuming log2 = 0.3010, log3 = 0.4771 and log7 = 0.8451, evaluate:
(i) \( \scriptsize \log_{10}21 \)
(ii) \( \scriptsize \log_{10}\sqrt{45} \)
(iii) \( \scriptsize \log_{10}12.25 \)