Example 1: = 0.244

1. Set the hair line over 4.1 on the LL₃ scale.
2. Under the hair line read off 0.244 on the LL₀₃ scale.

Example 2: = 1.071

1. Set the hair line over 0.9344 on the LL₀₁ scale.
2. Under the hair line read off 1.071 on the LL₁ scale as the answer.

Note

1. The reciprocal of any number from 0.000045 to 22,000 (10^-5 to 10⁵ for Slide Rules with extended scales) can be obtained by locating the number on the particular e^x or e^-x scale and reading off on the corresponding e^-x or e^x scale as in examples above.
2. Even for a Slide Rule with LL₀ and LL₀₀ scales, there is a gap between 0.9991 and 1.0009, but this range would not often be encountered. For number between 0.1 and 6, the LL scales give more accuracy the CI and D scales, while outside this range (i.e. greater than 6 and less then 0.1) the CI and D scales are more accurate. One of the greatest advantages of the LL scales for reciprocals, is that the decimal point is read directly off the scales.

Exercise 19(a)

1. =
2. =
3. =
4. =
5. =
6. =
7. =
8. 

Example: Find 2.5¹⁰, , , and . (Fig 19.2)

1. Set the hair line over 2.5 on the LL₂ scale.

Under the hair line red off –

2. 9,500 on the LL₃ scale as the value of 2.5¹⁰
3. 1.096 on the LL₁ scale as the value for .
4. 0.000105 on the LL₃ scale as the value for .
5. 0.9124 on the LL₀₁ scale as the value for

Further Examples:

For ‘a’ on the LL scale as shown, the other LL scales give –

1. 1.08¹⁰ =
2. 1.08¹⁰⁰ =
3. =
4. =
5. =
6. =
7. =
8. =
9. =
10. 0.65¹⁰ =
11. =
12. =
13. =
14. =
15. 1.7¹⁰ =
16. 0.97¹⁰⁰ =

Note:

1. There is no difficulty with locating decimal points, as all values are read off the particular LL scale.
2. Recall

The LL scales can be used to obtain a^N for a > 0 and N, any power (positive, negative or fractional). The procedure is as follows: For ‘a’ on the LL scale, the D scale give ‘ln a’ (see unit 18). If we multiply this value of ‘ln a’ on the D scale by ‘N’ (using the C and D scale) we then obtain ‘N ln a’ (i.e. ln a^N) on the D scale. If the multiplication was done using the left index of the C scale (see example 1 below), the value of ‘a^N’ is read off –

1. The original LL scale for 1 < N < 10.
2. The LL scale above the original for 10 < N < 100 (e.g. ‘a’ on the LL₂ then ‘a^N’ on the LL₃ or ‘a’ on LL₀₂ then ‘a^N’ on LL₀₃).
3. The LL scale below the orginal for 0.1 < N < 1. (e.g. ‘a’ on the LL₂ then ‘a^N’ on the LL₁ or ‘a’ on LL₀₂ then ‘a^N’ on LL₀₁).

In multiplying ‘N’ by ‘ln a’, on the C and D scales, if the right index of the C scale is used, the answers are found on the LL scale above the LL scales listed in the foregoing method. (see example 2).

This method can be extended to larger or smaller number values of N by moving two or more LL scales instead of one.

Example 1: Find 1.4^2.5,1.4²⁵, and 1.4^0.025

1. Set the hair line over 1.4 on the LL₂ scale.
2. Place the left index of the C scale under the hair line.
3. Reset the hair line over 2.5 on the C scale.

Under the hair line read off –

4. 2.32 on the LL₂ scale as the value for 1.4^2.5.
5. 4,500 on the LL₃ scale as the value for 1.4²⁵.
6. 1.0878 on the LL₁ scale as the value for 1.4^0.25.
7. 1.0084 on the LL₀ scale as the value for 1.4^0.025.

Example 2: Find 0.6^6.15, 0.6^0.615, 0.6^0.00615, 0.6^-6.15, 0.6^-0.0615

1. Set the hair line over 0.6 on the LL₂ scale.
2. Place the right index of the C scale under the hair line.
3. Reset the hair line over 0.615 on the C scale.

Under the hair line read off –

4. 0.043 on the LL₀₃ scale as the value for 0.6^6.15.
5. 0.73 on the LL₀₂ scale as the value for 0.6^0.615.
6. 0.99686 on the LL₀₀ scale as the value for 0.6^0.00615.
7. 23.3 on the LL₃ scale as the value for 0.6^-6.15.
8. 1.032 on the LL₁ scale as the value for 0.6^-0.0615.

Note:

1. 5.9⁹ = 5.9^4.5 x 5.9^4.5 = (5.9^4.5)². Use the LL scales to evaluate 5.9^4.5 and then square the results in the usual way.
2. 36.4⁵ = (3.64 x 10)⁵ = 3.64⁵ x 10⁵.
3. 1.52²⁹ = 1.52¹⁹ x 1.52¹⁰. Evaluate each separately and multiply together in the usual way or 1.52²⁹ = 1.52^14.5 x 1.52^14.5 = (1.52^14.5)²
4. 163^13.8 = (1.63 x 10²)^13.8 = 1.63^13.8 x 10^27.6 = 1.63^13.8 x 10^0.6 x 10²⁷. Evaluate the first term using the LL scales. To find the second term, use the L scale as 10^0.6 = x is equivalent to log₁₀x = 0.6 (i.e. find 0.6 on the LL scale and read off 3.98 on the D scale (or W’₂ scale for the 2/83N) as 10^0.6)
5. 0.0021^5.7 = (.21 x 10^-2)^5.7 = 0.21^5.7 x 10^-11.4 = 0.21^5.7 x 10^0.6 x 10^-12

1. 3.7^6.1 =
2. 3.7^0.61 =
3. 3.7^-0.61 =
4. 0.625³ =
5. 0.179^4.3 =
6. 0.02^1.7 =
7. 79² =
8. 70^0.14 =
9. 79^-0.002 =
10. 4,200^0.003 =
11. 0.9154.2 =
12. 22^-1.3 =
13. 44^-0.75 =
14. 1.01⁷⁰ =
15. 0.98³⁴ =
16. 14²² =
17. e¹⁴ =
18. 1,100^-4.8 =
19. 0.0057^8.3 =
20. 0.042^-19 =

Expressing we can the LL scale to obtain the root of any positive number. As previously with "powers" in 19.3, for ‘a’ on an LL scale the D scale gives ‘ln a’. If we divide this value of ‘ln a’ by ‘N’, we then obtain ‘’ (i.e. ) on the D scale. We must note the location of the decimal point for ‘’ so as to decide on which LL scale the value of ‘’ will be found. This is done according to the same rules used in 19.3.

Example 1 Find and

1. Set the hair line over 16 on the LL₃ scale.
2. Place the 3 of the C scale under the hair line.
3. Reset the hair line over the right index of the C scale.

(Note: and , thus we will find on the LL₂ scale and on the LL₁ scale.)

Under the hair line read off –

4. 2.52 on the LL₂ scale as the value for .
5. 1.0968 on the LL₁ scale as the value for

Example 2: Find and

1. Set the hair line over 420 on the LL₃ scale.
2. Place the 5 of the C scale under the hair line.
3. Reset the hair line over the left index of the C scale.

Under the hair line read off –

4. 3.35 on the LL₃ scale as the value for .
5. 0.298 on the LL₀₃ scale as the value for

Note: If we use the right index of the C scale when we divide by ‘N’ for 1 < N < 10 the value for ‘’ will be found on the LL scale below the original LL scale on which ‘a’ was located. While, if 10 < N < 100 the value for ‘’ will be found on the second LL scale below the original one.

If we use the left index of the C scale when we divide by ‘N’, for 1 < N < 10 the value for ‘’ will be located on the same LL scale on which ‘a’ was located. While, if 10 < N < 100 the value for ‘’ will be found on the LL scale below the original one.

Some special powers are given in the following table.

The LL scales can be used to obtain logarithms to any base, by placing the left or right index of the C scale over the base as found on the LL scale. Then for any number on an LL scale its logarithm to the chosen base is read off the C scale.

Example 1: Log₆23.5=1.765

1. Set the hair line over 6 on the LL₃ scale.
2. Place the left index of the C scale under the hair line.
3. Reset the hair line over 23.5 on the LL₃ scale.
4. Under the hair line read off 1.765 on the C scale as the answer.

Note:

1. We could have used the CF scale above, by placing the index of the CF scale under the hair line in step 2, and thus reading 1.765 off the CF scale in step 4.
2. For logarithms of other numbers to base 6, leave the slide as positioned in step 2 and reset the hair line over the number on an LL scale. (e.g. log₆2 = 0.3875 as 2 on the LL₂ scale give 0.3875 on the C scale.)

Example 2: Find log₂1.4, Log₂23 and log₂650.

1. Set the hair line over 2 on the LL₂ scale.
2. Place the right index of the C scale under the hair line.
3. Reset the hair line over 1.4 on the LL₂ scale and read off 0.485 on the C scale as the value for log₂1.4.
4. Reset the hair line over 23 on the LL₃ scale and read off 4.52 on the C scale as the value for log₂23.
5. Reset the hair line over 650 on the LL₃ scale and read off 9.34 on the C scale as the value for log₂650.

B. Solving exponential equations

Because Log_bN = x is equivalent to N = b^x the above problems are the same as solving an exponential equation for an unknown power. Example 1 could have been stated – Find x for 6x = 23.5. Thus to solve for an unknown power or exponent we could either express the equation in logarithmic form or leaving it as an exponential equation proceed as follows for N = b^x.

1. Set the hair line over the base ‘b’ on the appropriate LL scale.
2. Place the left or right index of the C scale under the hair line.
3. Reset the hair line over the number ‘N’ on its appropriate LL scale.
4. Under the hair line read off the value for x on the C scale and locate the decimal point according to the LL scales used.

Note:

1. To solve the equation for x, we follow the same method as outlined above, except in step 4 the value of x is read off the CI scale.
2. To solve the equation

Example:

1. Set the hair line over e on the LL₃ scale.
2. Place the 1.4 of the CIF scale under the hair line. (We can also use CI scale here if suitable).
3. Reset the hair line over 9 on the LL₃ scale.
4. Under the hair line read off 1.57 on the CF scale as the value for x.

Exercise 19(e)