Antil Log Tables from Log Tables

[Srinivasulu B]

For my children I wanted to have practice in using log table while
multiplying and dividing.
For this I have a book Clark's Tables (Science Data Book) by Orient Longmans
This has Log and Anti Log tables.
This is what I have been using in my school days.
Now this book is not easily available.
I wanted to take a hard copy of the these Log and anti log tables.
But I was not successful...as the paper is quite soiled.
Then I made search on the net.
I got a link.
http://www.sosmath.com/tables/logtable/logtable.html
Here I have log tables.
I also would like to have Antilog tables also...which is not available in
this site.
How can I generate Anti Log table using this table.
Any help?
seena

 

[smartin]

For my children I wanted to have practice in using log table while
multiplying and dividing.
For this I have a book Clark's Tables (Science Data Book) by Orient Longmans
This has Log and Anti Log tables.
This is what I have been using in my school days.
Now this book is not easily available.
I wanted to take a hard copy of the these Log and anti log tables.
But I was not successful...as the paper is quite soiled.
Then I made search on the net.
I got a link.
http://www.sosmath.com/tables/logtable/logtable.html
Here I have log tables.
I also would like to have Antilog tables also...which is not available in
this site.
How can I generate Anti Log table using this table.
Any help?
seena

Isn't an "antilog table" just a log table used "in reverse"?

Given

X = Antilog (0.301)

In other words,

X = 10^0.301

Or,

log(X) = 0.301

So, look for the closest match of log X = 0.301 in the log table, in
this case,

X = 2.00


Then the exercise can be extended:

Y = Antilog (1.301)

Well this means,

Y = 10^1.301

Or,

log(Y) = 1.301

But 1.301 is not in our log table.

However,

Y = 10^1.301 = 10^(1 + 0.301)

= 10^1 x 10^0.301

= 10 x 10^0.301

= 10 x Antilog (0.301)

Looking up Antilog 0.301 we have

= 10 x 2.00

Y = 20.0


Extending a little more, steps excluded:

Z = Antilog (-1.699)

Z = 10^(-1.699) = 10^(-2 + 0.301)

= 10^(-2) x 10^0.301

= 0.01 x Antilog(0.301)

= 0.01 x 2.00

= 0.0200


Hope this helps!

For more:
http://en.wikipedia.org/wiki/Logarithm

 

[Bernie Deitrick]

In cell A2, enter the formula
=(ROW()-ROW($A$2))/1000 + INT((COLUMN()-COLUMN($A$2) +1)/2)*0.1

Then in B2, enter the formula
=10^A2

Copy A2:B2, and paste in A3:B101.

Then copy A:B, and paste in C:T

Then copy everything, and paste special values over the formulas, and you have your table of values.

HTH,
Bernie
MS Excel MVP

 

[Srinivasulu B]

Thanks Berie,
I will try and get back to you
seena

 

[Srinivasulu B]

Excellent!!!
I can't thank you enough for this Berrie...really I cant't thank you enough.
Just unbelievable!
seena

 

[Bernie Deitrick]

Seena,

You're welcome, and thanks for letting me know that you got my solution to work.

Bernie
MS Excel MVP

 

[Srinivasulu B]

Berrie,
Now you have given me a fish.
If you can give me the angler rod...I will NOT bother you in future
If it is not too much of a botheration for you...canm you explain the
formula you have given
seena

 

[Bernie Deitrick]

Seena,

Sure.

The formula is this.

=(ROW()-ROW($A$2))/1000 + INT((COLUMN()-COLUMN($A$2) +1)/2)*0.1

I wanted to create ten columns of 100 values - in columns A, C, etc. that
increment down the column by .001.

So, to perform that increment of .001, I used (ROW()-ROW($A$2))/1000

ROW() with no argument returns the row number of the cell with the formula,
ROW($A$2) always returns 2, so I could have written

=(ROW()-2)/1000

if I were sure that the formula was going into cell A2, and that no new rows
would be inserted above A2.

Anyway, in cell A2, that returns (2-2)/1000 or 0.000. In A3, it returns
(3-2)/1000, or 0.001, and so on.

Then I wanted to add an additional .1 for every time that the formula was
copied to the next column (skipping one). So I used

INT((COLUMN()-COLUMN($A$2) +1)/2)*0.1

Just like ROW, COLUMN returns the column number. But since I wanted to skip
columns, I used /2 and INT to return 0 in column A, 1 in column C, etc.
(In reality, I didn't need to use the INT or the +1 - I just copied a
formula that I had used before for a different but similar purpose...)

I could have also used
(COLUMN()-COLUMN($A$2))*0.05
or even
(COLUMN()-1)*0.05
but I wanted to be clear that the increment was 0.1 (in case you wanted to
change to another increment) and to have the formula work when a new column
A was inserted. Anyways, in column C, this returns all returns .1, then in
E .2, etc.

Also, instead of that formula, we could have typed 0 in cell A2, then used
=A2+.001 in cell A3, and copied that down. Then in C2, we would have used
=A2+.1, and copied that down, and then over.... There are so many different
ways to approach problems in Excel, and I just chose one which I use
frequently and that I had ready to hand.

HTH,
Bernie
MS Excel MVP

 

[Srinivasulu B]

Thank you very much...for the detailed explanation ...BERNIE
I just noticed that I have been refering you as BERRIE but I should do it as
BERNIE
The "n" has ben mistaken as "r"

seena