Topic Content:
- Calculations using Logarithms
- Multiplication and Division
- Power & Roots
The basic principles of calculation using logarithms depend on the laws of indices.
Example 5.6.1 – Multiplication and Division:
Use tables to work out the following:
1. 26.52 × 9.184
2. 912.4 ÷ 94.35
Solution
1. 26.52 × 9.184
1st Method
Using the tables
i.e \( \scriptsize 10^{1.4235} \: \times \: 10^{0.9630} \)
= \( \scriptsize 10^{1.4235 \:+ \:0.9630} \rightarrow \left(x^a \: \times \: x^b = x^{a \: + \:b} \right) \)
= \( \scriptsize 10^{2.3865}\)
Using the anti-log \( \scriptsize \rightarrow 2.435 \)
Answer = 243.5
2nd Method
No | Log |
26.52 | 1.4235 |
9.184 | + 0.9630 |
2.3865 | |
Antliog of 0.3865 = 2.435 | |
102 × 2.435 | |
Ans = 243.5 |
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2. \( \scriptsize 912.4\: \div \: 53.55 \)
Using the tables
1st Method
i.e \( \scriptsize 10^{2.9602} \: \div \: 10^{1.7288} \)
= \( \scriptsize 10^{2.9602 \:- \: 1.7288} \rightarrow \left(x^a \: \div \: x^b = x^{a \: – \: b} \right) \)
= \( \scriptsize 10^{1.2314} \) Using the anti-log \( \scriptsize \rightarrow 1.704\)
Answer = 17.04
2nd Method
No | Log |
912.4 | 2.9602 |
53.55 | –1.7288 |
1.2314 | |
Antliog of 0.2314 = 1.704 | |
101 x 1.704 | |
Ans = 17.04 |
Power & Roots:
Example 5.6.2:
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 5.6.3:
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) |
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