Solving Logarithm with table.

[chijioke]

I need help with this:

Evaluate the following by the use of four-figure table.


$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$






$\displaystyle b. ~\frac{118.6}{34.95} - \frac{2837}{1964}$






$\displaystyle c. ~\frac{2.874^3+6}{2.874^3-6}$





$\displaystyle d. ~2.908^3 + \sqrt[5]{248.4}$



$\displaystyle e. ~\sqrt[3]{130-\sqrt[3]{130}}$






$\displaystyle f. ~11.65 × 91.21-11.65^2$




$\displaystyle g. ~\sqrt[5]{1.21^4+3.2^3}$


l am having difficulty when it comes to working with problems involving addition and subtractions.


I have no problem working with these ones below because the operations do not involve addition and subtraction. Please take a look at them:



$\displaystyle i. ~\sqrt{\frac{173.8×14.7^2}{2.61^3}}$



$\displaystyle ii. ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61}$



$\displaystyle iii. ~\sqrt[3]{\left(\frac{565×0.0536}{249.1}\right)^2}$


solutions


$\displaystyle i. ~\sqrt{\frac{173.8×14.7^2}{2.61^3}}$


$$\begin{array}{|l|r|} \text{No.} & \text{Log} \\ \hline 173.8 & 2.2400 \\ \hline 14.7^2 & 1.1673 × 2 \\ \hline ~ & = 2.3346 \\ \hline 173.8 × 14.7^2 & 2.2400 \\ ~ & + 2.3346 \\ ~ & \overline{4.5746} \\ \hline 2.61^3 & 0.4166 × 3 \\ \hline ~ & = 1.2499 \\ \hline \frac{173.8 × 14.7^2}{2.61^3} & 4.5746 \\ ~ & -1.2499 \\ ~ & =\\ \hline ~ & 3.3247 \\ \hline \sqrt{\frac{173.8 × 14.7^2}{2.61^3}} & 3.3247 ÷ 2 \\ ~ & = 1.6624 \\ ~ & \text{antilog of} \\ ~ & 1.6624 \bumpeq 45.96 \\ \end{array}$$
$\displaystyle \sqrt{\frac{173.8×14.7^2}{2.61^3}}=45.96$

$\displaystyle ii. ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61}$

$$\begin{array}{|l|r|r|} \text{No.} & \text{Log} & ~ \\ \hline 42.79 & 1.6313 & 1.61313 \\ 0.6154^\frac{1}{2} & \bar{1}.7892 ÷ 2 & ~ \\ ~ & (\bar{1} + 0.7892) ÷ 2 & ~ \\ ~ & (\bar{2} + 0.7892) ÷ 2 & + \\ ~ & (\bar{1} + 0.8946) & \bar{1}.8946 \\ ~ & ~ & \overline{1.5259} \\ 8.737 & 0.9413 & - \\ 25.61 & +1.4084 & 2.3497 \\ \hline 0.1505 & \bar{1}.1762 & ~ \\ \hline \end{array}$$
$\displaystyle ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61} = 0.1501$



$\displaystyle iii. ~\sqrt[3]{\left(\frac{565×0.0536}{249.1}\right)^2}$


$$\begin{array}{|l|r|} \text{Number} & \text{Logarithm} \\ \hline 565 & 2.7520 \\ 0.0536 & \bar{2}.7292 \\ 565 × 0.0536 & \overline{1.4812} \\ 249.1 & 2.3964 \\ \frac{565×0.0536}{249.1} & \overline{\bar{1}.0848} \\ \left(\frac{565×0.0536}{249.1}\right)^2 & \bar{1}.0848 × 2 = \bar{2}.1696 \\ \sqrt[3]{\left(\frac{565×0.0536}{249.1}\right)^2} & = \frac{1}{3}(\bar{2}.1696)=\frac{1}{3}(\bar{3} + 1.1696) \\ ~ & = \bar{1}.389~86 \\ ~ & \bumpeq{\bar{1}.3899} \end{array}$$
$\displaystyle ~\sqrt[3]{\left(\frac{565×0.0536}{249.1}\right)^2}$

$\displaystyle = \text{antilog of}~\bar{1}.3899$


$\displaystyle = 10^{-1} × 2.454$


$\displaystyle = 0.2454$


$\displaystyle = 0.245 (3 \text{s.f})$


I was able to solve i, ii and iii because there was no addition subtraction involved. Now let me take a few shots at some of the problems involving subtraction and addition.


$\displaystyle s1.~ \frac{2.8808^3-1}{2.8808^3+1}$



$\displaystyle s2.~\sqrt[3]{7.9^3 + 2.1^3}$



$\displaystyle s3.~\sqrt{61.73^2 - 38.27^2}$


$\displaystyle s4.~\sqrt[3]{9.58^3 - 7.63^3}$



Solutions

$\displaystyle s1.~ \frac{2.8808^3-1}{2.8808^3+1}$

Using the idea of difference and sum of two cubes at the numerator and denominator respectively to simplify first.


$\displaystyle \frac{2.8808^3-1}{2.8808^3+1}=$



$\displaystyle \frac{(2.8808-1)(2.8808^2+2.8808+1)}{(2.8808+1)(2.8808^2-2.8808+1)}=$



$\displaystyle \frac{(1.8808)(12.18)}{(3.8808)(6.418)}=$



$\displaystyle \frac{22.91}{24.91}$


Putting the simplified expression in a table we have

$$\begin{array}{|l|r|} \text{No.} & \text{Log} \\ \hline \frac{22.91}{24.91} & 1.3600 \\ & - \\ ~ & 1.3964 \\ \hline 0.9196 & \bar{1}.9636 \end{array}$$
Antilog of $\displaystyle \bar{1}.9636=0.9196$


$\displaystyle \frac{2.8808^3-1}{2.8808^3+1}=0.9196$

$\displaystyle s2.~\sqrt[3]{7.9^3 + 2.1^3}$

applying the sum of two cubes
$\displaystyle \sqrt[3]{7.9^3 + 2.1^3}=$
$\displaystyle \sqrt[3]{(7.9 + 2.1)(7.9^2-(7.9)(2.1)+2.1^2)}=$

$\displaystyle \sqrt[3]{(10)(62.41-16.59+4.41)}=$

$\displaystyle \sqrt[3]{(10)(50.23)}=$
$\displaystyle \sqrt[3]{502.3}=$
putting in table

$$\begin{array}{|l|r|} \text{No.} & \text{Log} \\ \hline \sqrt[3]{502.3} & 2.701 × \frac{1}{3} \\ & = 0.9003 \end{array}$$antilog of $\displaystyle 0.9003=7.949$
$\displaystyle \sqrt[3]{7.9^3 + 2.1^3}=7.949$

$\displaystyle s3.~\sqrt{61.73^2 - 38.27^2}$
applying the idea of difference of two squares:

$\displaystyle \sqrt{61.73^2 - 38.27^2}=$

$\displaystyle \sqrt{(61.73 + 38.27)(61.73 - 38.27)}=$

$\displaystyle \sqrt{(100)(23.46)}=$

$\displaystyle \sqrt{2346}$

Putting in table, we be have:
$$\begin{array}{|l|r|} \text{No.} & \text{Log} \\ \hline \sqrt{2346} & 3.3703 ÷ 2 \\ & = 1.6852 \end{array}$$antilog of 1.6852 = 48.44 so
$\displaystyle \sqrt{61.73^2 - 38.27^2}=48.44$

$\displaystyle s4.~\sqrt[3]{9.58^3 - 7.63^3}$
applying difference of two cubes, we have
$\displaystyle \sqrt[3]{9.58^3 - 7.63^3}=$

$\displaystyle \sqrt[3]{(9.58 - 7.63)(9.58^2 + (9.58)(7.63) + 7.63^2)}=$

$\displaystyle \sqrt[3]{(1.95)(91.7764 + 73.0954 + 58.2169)}=$

$\displaystyle \sqrt[3]{(1.95)(223.0887)}=$

$\displaystyle \sqrt[3]{435.0230}=$

putting in table, we have:

$$\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline \sqrt[3]{435.0230} & 2.6385 ÷ 3 \\ \hline 7.577 & = 0.8795 \\ \hline \end{array}$$
$\displaystyle \sqrt[3]{9.58^3 - 7.63^3}=7.577$

As I said earlier, I need help with a to g. I don't know how to go about them because of the minus - and plus + sign involved. Thank you.

 

[topsquark]

$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

l am having difficulty when it comes to working with problems involving addition and subtractions.

Who told you that you can? The issue is that $log(a + b) \neq log(a) + log(b)$. You can't take the log of each expression and add them.

What you might try is something like this: Let $x^2 = 3.025^3$ and $y^2 = \sqrt[4]{759.4}$. Then you have the difference of two squares and
$log_b(x^2 - y^2) = log_b \Big ( (x + y)(x - y) \Big ) = log_b(x + y) - log_b(x - y)$

The trouble with this is that x - y < 0 so the second log is not defined. Maybe difference of two cubes? You would have to experiment.

-Dan

 

[Dr.Peterson]

This is an excessively long post; I don't care to read through it all. But I'll comment on a couple points.

l am having difficulty when it comes to working with problems involving addition and subtractions.

In general, since logs don't work with addition, you use logs to find the addends, add them, and then use logs to do whatever else has to be done.

For example:

Evaluate the following by the use of four-figure table.

$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

First calculate 3.025^3 by taking the log, multiplying by 3, and taking the antilog.

Then calculate the 4th root of 759.4 by taking the log, dividing by 4, and taking the antilog.

Then subtract the numbers. This is all you can do, and is how we did it 50 years ago before calculators.

And the tricks you were shown still require you to do additions yourself; I don't see that they are any easier and more secure than just doing what it says.

$\displaystyle ii. ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61}$

$$\begin{array}{|l|r|r|} \text{No.} & \text{Log} & ~ \\ \hline 42.79 & 1.6313 & 1.61313 \\ 0.6154^\frac{1}{2} & \bar{1}.7892 ÷ 2 & ~ \\ ~ & (\bar{1} + 0.7892) ÷ 2 & ~ \\ ~ & (\bar{2} + 0.7892) ÷ 2 & + \\ ~ & (\bar{1} + 0.8946) & \bar{1}.8946 \\ ~ & ~ & \overline{1.5259} \\ 8.737 & 0.9413 & - \\ 25.61 & +1.4084 & 2.3497 \\ \hline 0.1505 & \bar{1}.1762 & ~ \\ \hline \end{array}$$
$\displaystyle ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61} = 0.1501$

The actual value, with a calculator, is 0.150019985..., so there is a small rounding error in your answer; for that reason I'll work this one with logs myself to check your work.

log(42.79) = 1.6313
log(0.6154) = -0.2108; -0.2108/2 = -0.1054
So log of numerator is 1.6313-0.1054 = 1.5259.

log(8.737) = 0.9414
log(25.61) = 1.4084
So log of denominator is 0.9414+1.4084 = 2.3498.

So log of quotient is 1.5259-2.3498 = -0.8239.

Antilog of -0.8239 is 0.1500.

Now I look at your work, and see, first, a typo on the first line; and then your log of 8.737 differs from mine. The actual log, from my calculator, is 0.94136233571176112795191207450327; so it looks like yours is not rounded up from 3 to 4. I can't tell whether that is the fault of your table or your work.

I have to say, it would be a lot less work on both your part and ours if you would just attach an image of your hand work. Trying to extract the right pieces from your Latex, given that the site doesn't let me Reply to a selected piece and retain the coding, is ridiculously hard.

 

[chijioke]

The actual value, with a calculator, is .150019985..., so there is a small rounding error in your answer; for that reason I'll work this one with logs myself to check your work.

log(42.79) = 1.6313
log(0.6154) = -0.2108; -0.2108/2 = -0.1054
So log of numerator is 1.6313-0.1054 = 1.5259.

log(8.737) = 0.9414
log(25.61) = 1.4084
So log of denominator is 0.9414+1.4084 = 2.3498.

So log of quotient is 1.5259-2.3498 = -0.8239.

Antilog of -0.8239 is 0.1500.

Now I look at your work, and see, first, a typo on the first line; and then your log of 8.737 differs from mine. The actual log, from my calculator, is 0.94136233571176112795191207450327; so it looks like yours is not rounded up from 3 to 4. I can't tell whether that is the fault of your table or your work.

I have to say, it would be a lot less work on both your part and ours if you would just attach an image of your hand work. Trying to extract the right pieces from your Latex, given that the site doesn't let me Reply to a selected piece and retain the coding, is ridiculously hard.

I quite agree with you that there are some errors in my work. Now let me make some corrections.


$\displaystyle ii. ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61}$



$$\begin{array}{|l|r|r|} \hline \text{No.} & \text{Log} & ~ \\ \hline 42.79 & 1.6313 & 1.6313\\ 0.6154^\frac{1}{2} & \bar{1}.7892 ÷ 2 & ~ \\ ~ & (\bar{1} + 0.7892) ÷ 2 & ~ \\ ~ & (\bar{2} + 0.7892) ÷ 2 & + \\ ~ & (\bar{1} + 0.8946) & \bar{1}.8946 \\ ~ & ~ & \overline{1.5259} \\ 8.737 & 0.9414 & - \\ 25.61 & +1.4084 & 2.3497 \\ \hline 0.1500 & 2.3498 & \bar{1}.1762 \\ \hline \end{array}$$


$\displaystyle ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61} = 0.1500$

Is that okay now?

For example:

First calculate 3.025^3 by taking the log, multiplying by 3, and taking the antilog.

Then calculate the 4th root of 759.4 by taking the log, dividing by 4, and taking the antilog.

Then subtract the numbers. This is all you can do, and is how we did it 50 years ago before calculators.

That looks to me like not doing what they asked- using table. To me, is as good as using calculator to solve and save myself the stress.

 

[JeffM]

I quite agree with you that there are some errors in my work. Now let me make some corrections.


$\displaystyle ii. ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61}$



$$\begin{array}{|l|r|r|} \hline \text{No.} & \text{Log} & ~ \\ \hline 42.79 & 1.6313 & 1.6313\\ 0.6154^\frac{1}{2} & \bar{1}.7892 ÷ 2 & ~ \\ ~ & (\bar{1} + 0.7892) ÷ 2 & ~ \\ ~ & (\bar{2} + 0.7892) ÷ 2 & + \\ ~ & (\bar{1} + 0.8946) & \bar{1}.8946 \\ ~ & ~ & \overline{1.5259} \\ 8.737 & 0.9414 & - \\ 25.61 & +1.4084 & 2.3497 \\ \hline 0.1500 & 2.3498 & \bar{1}.1762 \\ \hline \end{array}$$


$\displaystyle ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61} = 0.1500$

Is that okay now?


That looks to me like not doing what they asked- using table. To me, is as good as using calculator to solve and save myself the stress.

I have no idea why anyone would subject students to doing computations with tables. As Dr. Peterson said, we used to rely on the tables to do calculations that required high precision because scientific calculators and inexpensive computers were not available before the 1970’s and 1980’s repectively. For approximate work, we used slide rules. Tables and slide rules used to reduce labor and errors.

There is nothing intellectually interesting in using tables, and today they do not save labor or reduce errors. I suppose it might be possible to use them in a single lesson to elucidate how they work, but these exercises appear completely pointless. They are not designed to give practical pointers or to elucidate important concepts.

 

[Dr.Peterson]

I quite agree with you that there are some errors in my work. Now let me make some corrections.

$\displaystyle ~\frac{42.79×\sqrt{0.6154}}{8.737×25.61} = 0.1500$

Is that okay now?

Your new answer is, of course, correct.

That looks to me like not doing what they asked- using table. To me, is as good as using calculator to solve and save myself the stress.

If you seriously think that you were required to do all the work in these problems, including addition, using the logarithm table, please show me both the complete instructions for these problems, and an example you were given in which you need to add, to show how they teach it.

I previously said

And the tricks you were shown still require you to do additions yourself; I don't see that they are any easier and more secure than just doing what it says.

because my first impression was that you were showing examples you were given, since they are not the problems you were asking for help with.. But what you said was,

Now let me take a few shots at some of the problems involving subtraction and addition.

So, is this just your own idea, or something you were taught as the right thing to do?

If the idea of using methods like this,

$\displaystyle s4.~\sqrt[3]{9.58^3 - 7.63^3}$
applying difference of two cubes, we have
$\displaystyle \sqrt[3]{9.58^3 - 7.63^3}=$

$\displaystyle \sqrt[3]{(9.58 - 7.63)(9.58^2 + (9.58)(7.63) + 7.63^2)}=$

$\displaystyle \sqrt[3]{(1.95)(91.7764 + 73.0954 + 58.2169)}=$

$\displaystyle \sqrt[3]{(1.95)(223.0887)}=$

$\displaystyle \sqrt[3]{435.0230}=$

putting in table, we have:

$$\begin{array}{|l|r|} \hline \text{No.} & \text{Log} \\ \hline \sqrt[3]{435.0230} & 2.6385 ÷ 3 \\ \hline 7.577 & = 0.8795 \\ \hline \end{array}$$
$\displaystyle \sqrt[3]{9.58^3 - 7.63^3}=7.577$

is intended as a way to avoid addition, you do realize that you had to do more additions than if you just added the two cubes, don't you? Why did you think this was appropriate?

I have long been aware that log tables are still taught in India and some other places, and have generally avoided criticizing this practice to students, who have to do as they are told; but I have to agree with JeffM that it is not worth teaching. If they not only teach you to use tables, but give the impression that you must use logs even for addition, then they are doing far worse than I had assumed. Logarithms are (were) a tool for simplifying multiplications and powers, not something you can be forced to use even when it is impossible.

 

[JeffM]

I have long been aware that log tables are still taught in India and some other places, and have generally avoided criticizing this practice to students, who have to do as they are told; but I have to agree with JeffM that it is not worth teaching. If they not only teach you to use tables, but give the impression that you must use logs even for addition, then they are doing far worse than I had assumed. Logarithms are (were) a tool for simplifying multiplications and powers, not something you can be forced to use even when it is impossible.

I was not aware that log tables were still part of the curriculum in India.

I have, however, seen that students from India are often assigned problems that seem to have no purpose other than to address types of problems found in exams that seem not to have changed since Lord Curzon was Viceroy. Perhaps such problems discourage some students from using valuable educational resources in what is still a relatively poor country.

I fear, however, that they must drive some creative minds out of the educational process altogether. It has been quite a few decades since India had a tryst with destiny: perhaps its math students should be allowed to step out from the old to the new.

 

[Deleted member 4993]

I was not aware that log tables were still part of the curriculum in India.

I have, however, seen that students from India are often assigned problems that seem to have no purpose other than to address types of problems found in exams that seem not to have changed since Lord Curzon was Viceroy. Perhaps such problems discourage some students from using valuable educational resources in what is still a relatively poor country.

I fear, however, that they must drive some creative minds out of the educational process altogether. It has been quite a few decades since India had a tryst with destiny: perhaps its math students should be allowed to step out from the old to the new.

I have graduated from elementary-middle-high school (1965) and bachelor of technology (1970) from India. I was taught to use log and trig tables - was NEVER forced to use tables (due to over-population there was scarcity of logs and hence log tables). I got my first "slide-rule" (those are based on logs - as you know) in my 3 rd. year of Engineering school. I was NEVER restricted from using slide-rule.

I think there is a misunderstanding involved somewhere. Sometimes teachers avoid using "external help" to show-off their arithmetic muscles. While teaching in college, I used to add and multiply in class-room before the students could finish typing in the calculator. And sometimes while proctoring quizzes in class-room I would declare " -5 for anybody who touches calculator" (for example to convert 6 1/2 ft long rod to inches). My students used to like these challenges - their tryst with destiny.

However, I do not think I putout any creative flame, with these challenges (my students called these torture). I have many students pursuing higher "degrees" in universities - and many of those tell me that they miss my torture. No creative flame was put out by non-use of calculator......

 

[chijioke]

Your new answer is, of course, correct.

Thank you.

If you seriously think that you were required to do all the work in these problems, including addition, using the logarithm table, please show me both the complete instructions for these problems, and an example you were given in which you need to add, to show how they teach it.

This solving with table matter was something I was thought way back in secondry school then. I was just practicing to ensure I really gained mastery on it when I stumbled on the types that involve addition and subtraction.
Most of the examples were shown on mutiplication, division, roots and powers.
Only one example was shown on addition and subtraction, and that was the example where I used difference of two squares.
The difference and sum of two cubes were not given as examples for solving the problem. I just thought that if difference of two squares is used to simply before putting in table to do away with the addition and subtraction, difference and sum of two cubes can be used as well.

I previously said

because my first impression was that you were showing examples you were given, since they are not the problems you were asking for help with..

Yes. Your impression was correct. And again the rule is that I have to show my work, what I know about the problems, where I am having issue and how I desired to be helped to make it easier for volunteers in rendering assistance. That is why you saw all that examples.

So, is this just your own idea, or something you were taught as the right thing to do?

The use of difference and sum of two cubes was my own idea. And I can see that it really worked.

If the idea of using methods like this,

is intended as a way to avoid addition, you do realize that you had to do more additions than if you just added the two cubes, don't you? Why did you think this was appropriate

This was appropriate because in away the addition and subtraction will be eliminated as the work enters the table.

Logarithms are (were) a tool for simplifying multiplications and powers, not something you can be forced to use even when it is impossible.

Do you think it is really impossible? Remember topsquark made some suggestions in post 2 of which am yet to experiment.

Do you realize that you had to do more additions than if you just added the two cubes, don't you? Why did you think this was appropriate?

Yes. But that is what the instruction demands.

Why did you think this was appropriate?

At least in a way the addition and subtraction will be eliminated before the work enters the table.

I have long been aware that log tables are still taught in India and some other places, and have generally avoided criticizing this practice to students, who have to do as they are told; but I have to agree with JeffM that it is not worth teaching.

Yes. I thought the same. But all things being equal, I am just trying to gain mastery on the topic and ensure that am not leaving something behind.

 

[Dr.Peterson]

Yes. Your impression was correct. And again the rule is that I have to show my work, what I know about the problems, where I am having issue and how I desired to be helped to make it easier for volunteers in rendering assistance. That is why you saw all that examples.

The use of difference and sum of two cubes was my own idea. And I can see that it really worked.

These are contradictory statements. The "(first) impression" I referred to was that s1-s4 were examples you were GIVEN in a textbook or elsewhere, NOT your own work. I presume you are saying that only the one example you mention involved addition at all, and you chose to do these four problems the same way.

Were those 4 problems also from the same source as the others?

Only one example was shown on addition and subtraction, and that was the example where I used difference of two squares.
The difference and sum of two cubes were not given as examples for solving the problem. I just thought that if difference of two squares is used to simply before putting in table to do away with the addition and subtraction, difference and sum of two cubes can be used as well.

And again the rule is that I have to show my work

But that is what the instruction demands.

I asked you to show the actual instructions. You are certainly misunderstanding the requirement. Please comply with this request, and also tell us where the instructions come from -- is this preparation material for some particular exam, for example?

The fact that they used a difference of squares in one example (presumably one in which that actually saved a little work) does not imply that you must always use such methods.

I am just trying to gain mastery on the topic and ensure that am not leaving something behind.

You are going beyond any reasonable requirement. As I said, if you use log tables at all, they are a tool to make work easier (than doing it by hand) There is no reason to do things a harder way in order to use a tool.

 

[chijioke]

The "(first) impression" I referred to was that s1-s4 were examples you were GIVEN in a textbook or elsewhere, NOT your own work. I presume you are saying that only the one example you mention involved addition at all, and you chose to do these four problems the same way.

No, they're not examples given in a text book. They are my own work. At least, I have a clue about the problems involving addition and subtraction.
These are are actually problems from the text book I picked and solved to show what I know about the problems involving addition and subtraction.

Were those 4 problems also from the same source as the others?

I asked you to show the actual instructions.

Which one is the actual instruction again. There are no two actual instruction. The actual instruction is only one and that is

1. Evaluate the following by the use of four- figure tables.
2. Use Logarithm tables to evaluate
3. Evaluate, using Logarithm tables:

I uploaded the instruction verbatim if that will clarify some things. View attachment 33906

You are certainly misunderstanding the requirement. Please complying with this request, and also tell us where the instructions come from -- is this preparation material for some particular exam, for example?

I am just building myself. I am not preparing for any exam for now. I just picked some of my old text book, trying to help myself.

The fact that they used a difference of squares in one example (presumably one in which that actually saved a little work) does not imply that you must always use such methods.

Yes, of course. And that's the reason I came here to see if there are other similar methods.

You are going beyond any reasonable requirement. As I said, if you use log tables at all, they are a tool to make work easier (than doing it by hand) There is no reason to do things a harder way in order to use a tool.

What tool could that be? Calculator?

 

[Dr.Peterson]

These are are actually problems from the text book I picked and solved to show what I know about the problems involving addition and subtraction.

I just picked some of my old text book, trying to help myself.

Thanks: that's what we've wanted, information about the source, and the exact wording, to make sure this isn't for some exam that gives extreme restrictions, or problems that specify some extraordinary methods must be used.

Of course, as has been mentioned, if you aren't preparing for an exam that requires logs, then probably there is no benefit to using logs, as they are really a lot of extra work you don't need to do in today's world. If you just like the mental exercise, feel free to do so; but you can relax about what is required.

Which one is the actual instruction again. There are no two actual instruction. The actual instruction is only one and that is

1. Evaluate the following by the use of four- figure tables.
2. Use Logarithm tables to evaluate
3. Evaluate, using Logarithm tables:

In other words, the problems come from various exercises with slightly different wording, but say the same basic thing.

Here is what you need to know: "Evaluate by the use of four- figure tables", and "use Logarithm tables" don't mean that every step must be done with the tables; they simply mean, "use the tables where necessary". In using logs, it is assumed that you can add, and can do simple multiplications; you are asked to use logs to do the hard parts. (For example, when you multiply a log by 3 to cube something, you don't then find the logs of 3 and of the log itself and add them; you just multiply by 3 by hand. Not everything must be done by logs.)

When you see a trick that will make it easier, you can use it if you wish, but they are probably expecting you just to do the work as written. You haven't shown us the example you mentioned that showed using the difference of squares, but I am sure that example didn't say this is something you must do, and that it is part of what they mean by "use logs". They were probably saying, Look, here is an interesting trick you can sometimes use to change how you solve something.

What tool could that be? Calculator?

What I said was that logarithm tables are a tool. So when I say, "if you use log tables at all, they are a tool to make work easier (than doing it by hand). There is no reason to do things a harder way in order to use a tool", I am saying that when you are told to use logs, it means to use them as taught, where needed. It doesn't mean that you have to find the hardest possible way, or the most elaborate possible way, or to make sure that logs are the last thing you do. If someone told you to use a screwdriver to take a box apart, they don't mean you can only touch it with the screwdriver; they mean that the screwdriver will make the work easier. You're still allowed to hold it in your hands. They want to help you, not to impose restrictions on you.

The point is, you already know what you need to know. You don't need to ask about special ways to handle addition; you just add, because that's easy. You're making things hard for yourself unnecessarily.

 

[chijioke]

You haven't shown us the example you mentioned that showed using the difference of squares, but I am sure that example didn't say this is something you must do, and that it is part of what they mean by "use logs".

Just in case you insist on seeing examples. Here are examples verbatim.

 

[JeffM]

Well, this is EXACTLY what Dr. Peterson has been saying. They use logarithms when they need to.

They use a formula rather than logarithms to restate

$$x = 45.23^2 - 24.72^2 = (45.23 + 24.72)(43.23 - 24.72)$$
Next, they use basic arithmetic rather than logarithms to restate

$$x = (45.23 + 24.72)(45.23 - 24.72) = 69.95 \times 20.51$$
Now they use logarithms to avoid multiplication

$$x = 69.95 \times 20.51 \implies log(x) = \log(69.95) + \log(20.51) = 1.8448 + 1.3120 = 3.1568$$
Finally, rather than logarithms, they use antilogs, which are not even mentioned in the instruction, to get the answer

$$log(x) = 3.1568 \implies x = \log^{-1} (3.1568) = 1435.$$
Then they check using approximations. This was a highly practical exercise (before calculators) on using logarithms where needed to save work.

Of course, it is a method of no practical use today. I still have a book of tables, but my son, who is an engineer, has never needed one.

 

[chijioke]

I think since they could not give as many examples in solving:

$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

$\displaystyle b. ~\frac{118.6}{34.95} - \frac{2837}{1964}$

$\displaystyle c. ~\frac{2.874^3+6}{2.874^3-6}$

$\displaystyle d. ~2.908^3 + \sqrt[5]{248.4}$

$\displaystyle e. ~\sqrt[3]{130-\sqrt[3]{130}}$

$\displaystyle f. ~11.65 × 91.21-11.65^2$

$\displaystyle g. ~\sqrt[5]{1.21^4+3.2^3}$

I need to take a rest and maybe come back later. Thank you.

 

[chijioke]

Am back. I have not really giving up. I start with a.
$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

Solution
$\displaystyle ~ 3.025^3 - \sqrt[4]{759.4}=$


$\displaystyle ~ 3.025^3 - (759.4)^\frac{1}{4}=$


$\displaystyle ~ 3.025^3 - (7.59×10^2)^{\frac{1}{4}}=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^{2×\frac{1}{4}})=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^\frac{1}{2})=$

$\displaystyle ~ 3.025^3 - (\sqrt[4]{7.59} × \sqrt{10})=$

How do I simply further?

 

[topsquark]

Am back. I have not really giving up. I start with a.
$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

Solution
$\displaystyle ~ 3.025^3 - \sqrt[4]{759.4}=$


$\displaystyle ~ 3.025^3 - (759.4)^\frac{1}{4}=$


$\displaystyle ~ 3.025^3 - (7.59×10^2)^{\frac{1}{4}}=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^{2×\frac{1}{4}})=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^\frac{1}{2})=$

$\displaystyle ~ 3.025^3 - (\sqrt[4]{7.59} × \sqrt{10})=$

How do I simply further?

Without using a calculator that's as far as you can get.

-Dan

 

[JeffM]

Am back. I have not really giving up. I start with a.
$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

Solution
$\displaystyle ~ 3.025^3 - \sqrt[4]{759.4}=$


$\displaystyle ~ 3.025^3 - (759.4)^\frac{1}{4}=$


$\displaystyle ~ 3.025^3 - (7.59×10^2)^{\frac{1}{4}}=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^{2×\frac{1}{4}})=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^\frac{1}{2})=$

$\displaystyle ~ 3.025^3 - (\sqrt[4]{7.59} × \sqrt{10})=$

How do I simply further?

I do not see that $3.025^3 - ( \sqrt[4]{7.59} \times \sqrt{10})$ is

simpler than $3.025^3 - \sqrt[4]{759}$.

How do you define “simpler”?

The fact is that, if you want an approximate numerical value for either expression, you have to compute approximate numerical values for $3.025^3$ and $\sqrt[4]{759}$ or equivalently difficult expressions.

With modern technology, you simply use a calculator without any of these shenanigan.

Before hand calculators, you would have used logarithms and gone

$$\log (3.025^3) = 3 * \log (3.025).\\ log (\sqrt[4]{759}) = \dfrac{1}{4} \log (7 .59 * 10^2) = \dfrac{1}{4} * \{ \log (7.59) + 2\} = \dfrac{\log (7.59)}{4} + 0.5.$$
Then you would have used antilogarithms. And finally started in on the actual (approximate) subtraction.

It is a lot of work and unlikely to get an approximation as good as the calculator.

 

[Deleted member 4993]

Am back. I have not really giving up. I start with a.
$\displaystyle a.~ 3.025^3 - \sqrt[4]{759.4}$

Solution
$\displaystyle ~ 3.025^3 - \sqrt[4]{759.4}=$


$\displaystyle ~ 3.025^3 - (759.4)^\frac{1}{4}=$


$\displaystyle ~ 3.025^3 - (7.59×10^2)^{\frac{1}{4}}=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^{2×\frac{1}{4}})=$

$\displaystyle ~ 3.025^3 - (7.594^\frac{1}{4}×10^\frac{1}{2})=$

$\displaystyle ~ 3.025^3 - (\sqrt[4]{7.59} × \sqrt{10})=$

How do I simply further?

If I were forced do this with log tables, I would calculate 3.025^3 and (7.594)^(1/4) separately and subtract using fingers.

 

[chijioke]

Since there is no other way forward. I have to do what is advised.
$\displaystyle ~ 3.025^3 - \sqrt[4]{759.4}=$
Using log table
$\begin{array}{ll} \hline \text{No.} & \text{Log} \\ \hline \text{starting with minuend} & \\ 3.025^3 & 0.4807 × 3 \\ & = 1.4421 \\ \hline \text{antilog of}~1.4421 & \\ = 27.6758 & \\ \hline \text{working with subtrahend} & \\ \hline \sqrt[4]{759.4} & 2.8805 × \frac{1}{4} \\ & = 0.7201 \\ \hline \text{antilog of}~ 0.7201 & \\ = 5.2493 & \\ \hline \text{doing what they said, i.e subtracting} & \\ \hline 27.6758 & \\ - & \\ 5.2493 & \\ \hline 22.4265 & \\ \bumpeq{22.43} & \\ \hline \end{array}$