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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6 Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3 Topics
  3. Number Base System III | Week 3
    2 Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2 Topics
  5. Modular Arithmetic II | Week 5
    2 Topics
  6. Modular Arithmetic III | Week 6
    3 Topics
    |
    1 Quiz
  7. Indices I | Week 7
    3 Topics
  8. Indices II | Week 8
    1 Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3 Topics
  10. Logarithms II | Week 10
    4 Topics
    |
    1 Quiz



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Example 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

NoLog
\( \scriptsize \left (3.9562 \right)^3 \)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

NoLog
\( \scriptsize \sqrt [6] { 68.15} \)
\( \scriptsize \left( 68.15 \right)^{\frac{1}{6}} \)1.8335 ÷ 6
= 0.3056
Antilog of 0.3056
= 2.021
100 × 2.021
Ans = 2.021

Example 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

NoLog
\( \scriptsize(18.6)^2 \)1.2695 × 2 =2.5390
9.760.9894 +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

NoLog
43.121.6347
4.08+0.6107
2.24542.2454
3.401+0.5316
2.1840.3393
0.87090.8709
\(\scriptsize \left[ 1.3745 \right]^3\) = 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) \( \scriptsize \log_y \normalsize \frac{1}{16} = – \frac{1}{2} \)

(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) \( \scriptsize \log_{10}12.25 \)

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