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

Srinivasulu B wrote:
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 mo
http://en.wikipedia.org/wiki/Logarithm

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" wrote in message
...
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

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



"Bernie Deitrick" <deitbe @ consumer dot org wrote in message
...
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" wrote in message
...
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

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


"Bernie Deitrick" <deitbe @ consumer dot org wrote in message
...
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" wrote in message
...
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

Seena,

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

Bernie
MS Excel MVP


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

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" <deitbe @ consumer dot org wrote in message
...
Seena,

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

Bernie
MS Excel MVP


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

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" wrote in message
...
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" <deitbe @ consumer dot org wrote in message
...
Seena,

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

Bernie
MS Excel MVP


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

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

"Bernie Deitrick" <deitbe @ consumer dot org wrote in message
...
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" wrote in message
...
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" <deitbe @ consumer dot org wrote in message
...
Seena,

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

Bernie
MS Excel MVP


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