Logarithms & antilogarithms + How to find Logs & antilogs using Log Table



"Study is about taking the concepts & Laws of the subject/ topic"

Hello Guys

Thank you for coming here. I've recently created a video about the concept of Logs & Antilogs, with a few more things than the article itself. That way (with video), you can learn even better & interactively as it features my own voice commentary. Check that out at:

youtube.com/watch?v=-0iKT4bdNuU

Please visit the above link, to like & Subscribe on YouTube, if you liked it. :) The embedded video:



I hope you like it

I've noticed that some people have confusion about logarithms (& anti-logarithms) & find it tough to find. Like anything else it's not very difficult thing. So here are common, clear & understandable notes on the same. Here's a graph for normal Logarithmic Function:


Photo By Ellywa from nl, From Wikimedia Commons, licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.

So today I'll try to explain some general principles about logarithm & its use, I'll try to keep this guide precise and as clear as possible.

This guide will be especially helpful to students who have recently joined class 11, especially in India.

I'll also give an example to learn the use / application of logarithms, so that you may learn the power of it.

In short, it is everything you need to know about logarithms & antilogarithms (but not as much as wiki!) :P
BASE TO & VALUE

Logarithm is not a very tough thing to understand (like anything else!).

Before learning logarithms we'll need to have information about "Base to" & "Value", terms that I've named. :P lol!


Logarithm & Antilogarithms deal with 3 kinds of numbers, I call them "Base to" & "Value" & "answer". "Base to" refers to the basic sign convention or number used.

"Value" refers to the logarithmic value used for a particular "Base to".

For example,
logbv = a
Here b= Base To, v= value, a =  answer that we get.

"Base to" remains same mostly, but "value" mostly changes.

The most common "Base To" are:
1. e (exponent), which is like π (pie) but doesn't have a constant value. It varies as ex where 'x' is any rational number.

2. 10, which is the number 10 we use.

Similar to Logarithms, Antilogarithms also have base & value, they just give the value opposite to Logarithms, i.e. whose log has been calculated (You'll learn calculation later in this article.)

WHAT ARE LOGARITHMS & ANTILOGARITHMS

Photo By ddpavumba / Freedigitalphotos.net

So now that we know "base to" & "value", we can see logarithms.

Take an example, someone asks you to calculate:

Antilog(al)103, then it'll be equal to:


Antilog(al)103= 103

Similarly,
Antilog(al)e4= e4.

Now Logarithm is the number associated with the opposite of antilogarithm, where "Base to" remains the same & vice versa. For example,

Comparing with he above examples,

Logarithm(log)10103 = 3

&

Logarithm(log)ee4 = 4.

PROPERTIES

So now you'll use the properties of logarithms (logs), to use its power.
Some properties of Logarithms are:

Note: These apply for not only 3 & 4, but for all rational numbers.

1. logarithm(log) (3*4) = Logarithm(log) 3 + Logarithm(log) 4, where "base to" remains the same on both sides of equation & can have any rational value.

2. Log(3/4) = Log 3 - Log 4, where "base to" remains the same on both sides of equation & can have any rational value.

3. Log(34) = 4*Log 3, where "base to" remains the same on both sides of equation & can have any rational value.

Hint: You can use the converse of equations too.
Note: These do not apply antilogarithms.

CALCULATIONS USING LOGARITHMIC TABLE (Log Table)

Now that we know logarithmic properties, well done if you've understood them, so let's get started with the use of log tables.

Log tables use Log 10 v, so I'll not be writing "Base to" here, i.e. they'll give you base to 10 log's answer.

The examples that I'll use below will give you complete clarity about the topic, so they are golden examples & must be seen.

Calculating logs

Learn one thing that while calculating Logs (not Antilogs!), we use scientific notation, that is decimal is after one number only. But we get answer in non-scientific notation.

Consider:

EXAMPLE 1. Log (0.8372) {Base to 10 remember!}

we'll write it in scientific notation:

= Log (8.372*10-1).

Now we'll use property of logs.

= -1 + Log (8.372)

------------------------------------------------------------------------------------------------------------

Now we'll find the Log (8.372), using log table. The procedure will be same for finding values (in table) of antilogs, tangents, sines, etc. The laws/properties will be different though.




In the table, first find 83 (for above example), in first types of column then find the value under 7 for 83 in second kind of columns & then find the value of 2 for 83, in the third kind of columns.


Now add all the 3 values, you'll get required logarithm base to 10.


9227 + 1

log (8.372) = 0.9228


------------------------------------------------------------------------------------------------------------
Now in the above example

log (0.8372) = -1 + Log (8.372)
= -1 + 0.9228
= -0.0772

What if we want the value Log e  , instead of Log 10  , using Logarithm Table.

So learn a rule,
Log e = 2.303 Log 10

Calculating Antilogs

Now I'll teach you antilogs, with another golden example.

EXAMPLE 2. Calculate Antilog (86.654), using Anti-logarithmic Table.

Now to do this, 

Antilog (86.654) = 1086.654
But is it possible to calculate it directly?
No. Of-course not.

Therefore, use this method:


Note: In Antilogs, we use non-scientific notation, but get the answer in scientific notation.

Note: Tables give "Base to" 10 remember!

Al (86.654) = 1086 * Al (0.654)

Now use the anti-logarithmic table in the same way as taught in example 1.

The answer is: 4.508 * 1086


Calculating Antilogs (Special case)

Did you know that we can also calculate Antilogs of negative numbers, here's another golden example to teach you the same.

EXAMPLE 3. Antilog ( -8.654), using Anti-logarithmic table.

Now for negative numbers, Use the whole number next to the given number, add & subtract it, a mathematical trick.

Al ( -8.654) = Al ( +9 -9 -8.654)

Now +9 -8.654 gives 0.346

= Al (-9 +0.346)

= Al ( 0.346) * 10-9

Now from Anti-logarithmic table calculate Al (0.346).

Therefore, answer is: 2.218 * 10-9

SOME IMPORTANT POINTS

So here I have some necessary golden (or silver :P) points, which are actually sometimes confused.
I've also summed up the important points mentioned earlier, so that you may not need to list them separately.
1. Tables ( Both Logs & Antilogs) have base to 10.

2. If we use loge , it is called natural Log (or ln)


3. Log e ( ) = 2.303 Log 10 ( ) = Ln ( )

4. Procedure followed to get the values of logs, antilogs, sines, tangents, cotangets, etc. are same in a typical Logarithmic book

5. Logarithms can't be negative, whereas Antilogarithms can have negative values.
And generally, the opposite is true for their answers. Logs can have negative answers & Antilogs can't.

6. While calculating Logs (not Antilogs!), we use scientific notation, that is decimal is after one number only. But we get answer in non-scientific notation.

7. In Antilogs, we use non-scientific notation, but get the answer in scientific notation.


8. Properties of Logs do not apply antilogarithms.

9. General Form of Logs:

logbv = a
Here b= Base To, v= value, a =  answer that we get.

10. Log 1 with any "Base to" is ZERO 0.


Logxx = 1, Here x can be 10, e or any other natural number.


PRACTICAL APPLICATIONS

So here I'll give a practical example or application of logarithms, so that you may be able to witness the power of logs.


EXAMPLE 4. Evaluate 245 x 35


Applying logarithms gives
log (245 x 35) = log 245 + log 35
= log (2.45 x 102) + log (3.5 x 101)
= log 2.45 + log 102 + log 3.5 + log 101
= 0.3892 + 2 + 0.5441 + 1
= 3.9333

Therefore, 245 x 35= al(3.9333)=8576

Actually LHS 245 x 35= 8575
which is close enough! Isn't it?

And if you have log book with you and no calculator, you won't need to calculate!

Note: I may not hold any responsibility, about the exact information, you can confirm it from anywhere else. This is a thesis.

Please reply with your comments. Even a "Thank You" will be sufficient for me to know that I was able to help you. You can also give your queries or reviews.
Photo By Boians Cho Joo Young / Freedigitalphotos.net

Hope You learned & understood the above & got clarity about the topic.


ENJOY!!!

Please Ask your question/Give your opinion, here, I love to read User Comments.

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210 comments:

  1. Amazing explanation.....i have thou roughly learned logs and antilogs now....i would rate a 5-Star to this article!..

    ReplyDelete
    Replies
    1. Thank you very much, I'm happy you liked the above article & also, happy to help!

      Delete
    2. How would you know...when to use logs or anti-logs

      Delete
    3. Hey, nice question.

      It depends on the question or formula you have, like in some, you're given logs & when you need value inside log, like log(x)=0.8, antilogs become handy & can be used to calculate x.

      Delete
    4. Amazing .... I loved this ..... this helped me a lot... thanks and I would surely rate this 5-star

      Delete
    5. Really glad to be able to help.

      Delete
    6. But how do we find logs of numbers
      which have a zero before decimal? For Example: log (0.05)

      Delete
    7. It's helpful to learn logs but antilogs are not that clear

      Delete
    8. though the logs were properly explained , I couldn't understand the anti-logs at all, U didn't even use the anti-log table to explain it . In brief the explanation of ant-logs sucked , waste of time.

      Delete
    9. I couldn't understood this topic in school but with the help of your article now I'm well-versed in it and can explain others.Thanks a lot

      Delete
  2. Replies
    1. how can we find the log of number whose digits are less than 4 like 90.1

      Delete
  3. yes!!! here the concept of log and antilog has been explained very well.thanks a lot from my side

    ReplyDelete
  4. nice explanation

    ReplyDelete
  5. Wow!! Better explained than my chemistry teacher!! Thanks

    ReplyDelete
    Replies
    1. hehe!!! :D You're welcome, I'm happy, you liked it.

      Delete
  6. THANK YOU VERY MUCH!!!!!!!!!!!!!!!!
    U SAVED MY CHEMISTRY EXAM........:D;D

    ReplyDelete
    Replies
    1. :) You're Welcome, I'm Blissful to help you

      Delete
  7. I was so worried about my physics exam....but now i am confident enough to score good marks...Thanks for your help... :)

    ReplyDelete
    Replies
    1. You're Welcome... All the best for your Exam. :)

      Delete
  8. thankyou so much for your help....

    ReplyDelete
  9. well done bro... :) a big thank u..

    ReplyDelete
    Replies
    1. Thank you too, for appreciation & You're welcome. :)

      Delete
  10. Appreciable job done for the sake of students preparing for their exams.
    Good effort. Keep contributing. Jai Hind
    Sanjay Kumar

    ReplyDelete
  11. helpful I it helped me a lot

    ReplyDelete
  12. can you tell how to find antilog of -0.9876

    ReplyDelete
    Replies
    1. Hi!

      Yes, Antilog of -0.9876 can be calculated in a similar fashion, as shown by Example 3 of this Article.

      I'm assuming that it's Antilog "base to" 10. If one has a scientific calculator, the answer may be calculated by the command 10^-0.9876 (raised to the power).

      To find it with Logarithmic table, do as follows (Remember it's Antilog base to 10):

      al(-0.9876)
      = al(+1-1-0.9876)
      = al(0.0124-1)
      = al(0.0124) * 10^-1

      Now consult a correct Antilog Table to find out al(0.0124). The method has been shown in the article.

      = 1.029 * 10^-1
      = 0.1029

      which is the answer :)

      Delete
    2. al(0.0124) is either 1028 or 1330

      Delete
  13. thank u for giving such a clear explanation

    ReplyDelete
    Replies
    1. Sir i am thankful for ur genius explanations and examples...regards

      Delete
    2. Thank u so much
      Explained better than chem teachers
      Can u pl tell me the antilog of 10^6

      Delete
  14. Replies
    1. Hey!

      I'm considering "base to" 10:

      log(0.1) = log(10^-1)
      = -1*log10
      = -1 (remember base to 10).

      which is the answer. :)

      Delete
  15. How to calculate log5-log3?

    ReplyDelete
    Replies
    1. Hi!
      I think it's a good question, as division by 3 leads to an ongoing infinite answer, but still we can round it off to 3 decimal places in scientific notation, which will only generate a slight error which MAY BE ACCEPTABLE . There can be more than 1 way to solve this question, three of which are:

      (I think all the ways might be acceptable. Although if your examination is very-very strict, ask your teacher first)

      1. log5 - log3 = -(log3 - log5)
      = -[log(3/5)]
      = -log(0.6)
      = -(-0.2218)
      = 0.2218

      2. We can calculate both the logs (log5 and log3) separately, to still get the answer:

      log5 - log3 = 0.6989 - 0.4771
      = 0.2218

      3. We can directly use the property of logarithm (without negative sign):

      log5 - log3 = log(5/3)
      = log(1.667) Rounded off.
      = 0.2219

      OR some Log books might give 0.222 as the answer, which is still acceptable.

      As you can see, even in case 3, there's not much difference in the answer either, so it's also acceptable.

      I hope you found your answer!

      Delete
  16. How can we calculate antilog of 1.9543

    ReplyDelete
    Replies
    1. Hi!
      It can be found using the methods given in this article only (I'm considering base to 10 here):

      Al (1.9543) = 10^1 * Al (0.9543)

      (Now consult an Antilogarithm table)

      = 10 * 9.001
      =90.01

      which is the answer.

      Delete
  17. tysm ur blog is very helpful 2 me

    ReplyDelete
  18. thank uuuuu vvvvvvvvvvery much.......................

    ReplyDelete
  19. Thnq u vry much ........it is vry helpful

    ReplyDelete
    Replies
    1. Hii its too esy n useful thnks wil u pls xplain characterstick n mantess

      Delete
    2. Hi Santosh!

      It's nice to see, you found the post useful. I've myself never put much stress on Characteristic & Mantissa, while dealing with Logarithms, because I've never felt a need to use them so currently I don't know much about them. So "Can no Explain" ;) for now. I remember that Logarithmic book (the one with Log Tables) had it, so maybe you could try reading it from there or some other place.

      Delete
  20. thank u soooo much.............its really helpfull ...................better than d teachers explaination..........................

    ReplyDelete
  21. I would like to know how to take antilog of numbers between 0 and 1. For example antilog(.0237). Would u please help me out.

    ReplyDelete
    Replies
    1. Hi Kailash!

      Actually the way I used here, actually uses non-scientific notation only (0.xxxx), to calculate Antilogs, the answer would then be in scientific notation. For Example, to calculate Antilog(0.0237) [Pick up a Log Book & solve with me side-by-side] :-

      Search for 0.02 int he rows of Antilog Table, then match it with the column 3 (because of the number 3 in the question), we'll get:

      1054

      For the final number, 7, we need to see the Mean Difference columns, where 7 for 0.02 would represent 2, add 2 to the 1054:

      1056

      Finally write the answer as (a rule):

      1.056, which is the required Antilog!

      I wish you're able to understand this. If not, comment again with your query.

      Delete
  22. I got a question which said find (1.056)^(1/3). Here's what I did Let x be the answer, so log x = 1/3log(1.056).
    I found log1.056 to be .0237 using log table
    So log x = .0079
    Now x = antilog(.0079)
    I found antilog to be 1.018
    This is not the exact answer as 1.018^3 is only 1.054977...
    The question asked me to round of the answer here x to the forth decimal place.
    But I was able to find only three decimal places, is there any way to find the 4th and 5th decimal place and round it to the 4th decimal. If not how do I proceed ???

    ReplyDelete
    Replies
    1. Actually this is the limitation of logarithmic tables. I don't think, there's a direct way to get to the fourth decimal place (Atleast not I know about, maybe there could be a way!). I think, when the examiner would put this kinda question, he'll know this limitation of logarithmic tables. Maybe 1.0180 could be the answer, but I'm not sure! If a teacher asked you this question, you should try consulting him/her.

      Delete
    2. Just use the method as mentioned in the post pal!

      Delete
  23. Thank you for helping me out. My of them in my class weren't sure on how to take log and antilog. So they did the other question in option. So I couldn't ask anyone in my class. You helped me a lot. Thank you again

    ReplyDelete
  24. Replies
    1. -log (1.41 x 10^-3)
      = - {log1.41 + log(10^-3)}
      = - {log1.41 - 3}
      = - { -2.85}
      = 2.85

      which should be the answer!

      Delete
  25. Thank you so much
    It is really helpful

    ReplyDelete
  26. thnks for sharing wonderful knowledge....it helped me a lot

    ReplyDelete
  27. Will u pls tell me how to calculate 1.079/0.0665?

    ReplyDelete
    Replies
    1. Good question! If you want to do this, using logarithms & antilogarithms:

      let x=1.079/0.0665

      Take log on both sides (base to 10):

      log x = log (1.079/0.0665)
      = log 1.079 - log 0.0665 --------------------1

      Now use a log table to calculate R.H.S. :
      log 1.079 is 0.033

      For Log 0.0665, write it as log (6.65 * 10^-2)
      which becomes "-2 + log(6.65)"
      or "-2 + 0.8228" (from log table and the method in the article).
      which gives -1.1772

      Now put the calculated values in equation 1:
      log x = 0.033 - (- 1.1772) {which is actually sum}
      = 1.21

      Now take antilog on both sides, to get:
      x = al(1.21)
      = 16.22 which is the answer approximately {from antilog table & method mentioned in the article}

      I hope you got it!

      Delete
  28. bhai tu to iit mein top krega :P

    ReplyDelete
  29. explanation of finding logarithm is nice. But i now also have some confusion in anti-logarithm. if there was some more examples, it would be fine....

    ReplyDelete
    Replies
    1. Thanx... I think the last example says it all. Antilog & Log of a number have METAPHORICALLY similar relation as multiplication & division have (not same rules). They're the opposites.

      If y = Log x
      then, Al y = x

      You may consult a Log Book, or class 11 & 12 books / above standard books, which give a lot of good examples.

      Delete
  30. Got clarity !!!
    But 1 thing that i don't know how to observe or see or find log value in log table........so can u pls provide above said tutorials...

    And can i get in touch with you through fb or twitter or skype handle....
    Thanks

    ReplyDelete
    Replies
    1. Hi Aditya

      I'm happy you liked it. You've already gone through a large part of logarithms. For Log Table, like the example of 8.372 in the article:

      First find 83 (as they're first 2 digits in the example), on the left hand side of a log table. Now, matching with the row of 83, find the digit 7 (the next one in example) as a column header and finally find 2 as a column header again matching the row of 83.

      Add both the found values as:
      9227 + 1 = 9228

      And finally write the answer as notation:
      0.9228

      I hope it's understandable now. If not, you may ask again. To connect with me on Facebook, I've sent you my ID as a message on your Google Plus. I'll be glad.

      Delete
  31. I knew it......
    That u would give example of like that only means that why had u taken an example of 1 digit i.e 8 ..... I want an explanation of 3 digit , wait, take log108 as example ..
    ...i am not able to solve my questions .... In june my exams are waiting for me!
    Just tell me in brief ... don't bother :)

    ReplyDelete
    Replies
    1. Hi Aditya

      I just used that number for instance. I'm not sure what you mean by 3 digit, could you tell me a number so that I can find its logarithm & show. Anyways here are a couple of examples:

      Log 8 (base to 10)
      = log 8.000

      So first find 80 in rows (left hand side), then under zero column find the answer, there's no need to add anything as the next number is zero. So it'll be:

      =0.9031, which is the answer.

      Taking a completely different example now. Consult a complete logarithm table for that.
      Let's find log (4.32).

      So finding 43 & 2 column (no need to add again), we have:

      0.6355 as the answer.
      I hope it's understandable now. If not, you may ask again.

      Delete
  32. thank you sir it was like my friend is teaching me ...very nice explination

    ReplyDelete
    Replies
    1. O! I'm very happy about that, you're welcome.

      Delete
  33. Not much helpful.

    ReplyDelete
    Replies
    1. Everyone has his/her own perspective & I respect that. Maybe it's because you already knew a lot about logs & antilogs.

      Delete
  34. sir ,
    can you give me a solution of e to the power -1.7925 by log method it is very helpful for me sir
    thanks

    ReplyDelete
    Replies
    1. Sure.

      Let e^-1.7925 = x.

      It can be written as:
      Aln(-1.7925) = x {base to e}

      Take Log base to e on both sides:
      -1.7925 = ln(x) {Natural log & base to e is same}

      OR:
      -1.7925 = 2.303 log(x) {base to 10}

      OR:
      log x = -0.7783 {base to 10}

      OR:
      x = Al(-0.7783) {base to 10}

      And finally:
      x = 0.1666, which is the answer.

      I hope you got it :)

      Delete
  35. Puneeth Deraje1 March 2015 at 20:57

    I had learnt log and antilog by trial and error method and so some things were a bit unclear but ur blog made it crystal clear.
    Thank you

    ReplyDelete
    Replies
    1. Really Glad to be able to make your points clear.

      Delete
  36. i am not able to see the log table properly, plz tell me easy way to find the log of the number......plz plz plz........

    ReplyDelete
    Replies
    1. Sure. All you need to do is find the only answer related to your number in the log table. Like to get Log(8.372) :

      Just find 2 digits (83 here) on the left hand side of the table.

      Then find the next digit (7 here), on the top. Match both the row & column to get your number. This is approximate answer,9227

      To find exact one, simply search for 3rd number (2 here) on the right & add that number to the answer just obatined.

      You should get 9228 as the result, which should be written as 0.9228 as the final answer.

      Best of luck.

      Delete
  37. thankx man....... well explained

    ReplyDelete
  38. thanks.... 🎓 🎓 🎓 🎓

    ReplyDelete
  39. Thank you..this Post was very Useful...
    But i would just suggest that....you can mention in the intro part that...Log to the Base 10 is 1!!! and e to the power log x is x!!!!
    Just to be clear....
    But everything was fine...
    Good!!!!

    ReplyDelete
  40. thankyou........ becoz of u im now abl to understand logaritms.. .... ... thankyousoooo much

    ReplyDelete
  41. THANK YOU! Finally I understand, the day before my board exam :p

    ReplyDelete
    Replies
    1. Hehe same here!!!

      Delete
    2. hehe that's the spirit :D All the best!

      Delete
  42. Thank you so so much! Boards from tomorrow and this was hugely helpful :)

    ReplyDelete
  43. While taking log, we got 9227 from the table, why did you make it 0.9227 when adding?

    ReplyDelete
    Replies
    1. PLS CAN ANY1 ANSWER THIS^^^^^^^^!!!

      Delete
    2. Well, when I was learning myself, I got quite confused about logs. After which I thought of a rule:
      "While calculating Logs (not Antilogs!), we use scientific notation, that is decimal is after one number only. But we get answer in non-scientific notation."

      It's working like a charm so far. You can simply learn that. I hope it's understandable.

      Delete
  44. Thank u Mohak Arora..... very well explained.....not got such simple explanation anywhere else.....!!!
    Glad dat u helped me out...!

    ReplyDelete
    Replies
    1. Thank you so much for appreciation. You are welcome.

      Delete
  45. Well Done; a great job. pls solve foll:

    0.9876x(16.42)2 (2 is exponent) divided by (4.567)1/3 (1/3 is exponent)

    ReplyDelete
    Replies
    1. Sure.

      If you want to solve both numerator & denominator by logs,
      Let [0.9876*(16.42)^2] / [(4.567)^1/3] = x

      Take Log (base to 10) on both sides:
      Log [0.9876*(16.42)^2] / [(4.567)^1/3] = Log x

      Use the properties of Logarithm:
      Log (0.9876) + 2 Log(16.42) - 1/3 Log(4.567) = Log x

      Consult a Logarithmic table to calculate the values.
      -0.0054 + 2 * 1.215 - 1/3 * 0.6596 = Log x

      Solve the equation:
      Log x = 2.205 (approximately)

      Take antilog on both sides:
      x = Al (2.205)

      which gives 160.3 as the answer.

      I hope you got it.

      Delete
  46. I find your this post much helpful ,although i know alot about logs and antilogs ... But i had gone through this bcuz i am good not the best O:) and i want to be the best atleast in logs and antilogs.... So i want u to clear just one doubt i.e. How to solve log108 and log 1210 .... Plz do solve this .
    And do give the fb or skype id or your twitter or instagram handle..
    I am a b.com hons student and i wanna top in maths .. I had got last year 99 marks out of 100 in 12 th class ... So plz reply fast

    ReplyDelete
    Replies
    1. Sure. I was busy, so I couldn't reply earlier, my bad, but here is the answer now. I'm considering Log base 10 only:

      log108 can be written as:
      Log(1.08 * 10^2)

      Use property of logs, it becomes:
      2 + Log(1.08)

      which gives:
      2 + 0.0334 = 2.0334, the answer

      Similarly for log 1210,
      log (1.21 * 10 ^3)

      = 3 + Log (1.21)
      = 3 + 0.0828
      = 3.0828 is the answer.

      I would also like to contact you, check out my facebook page at:
      on.fb.me/1CLXKTd

      There you can message me & I may add you as Friend.

      Delete
  47. Veer how we calculate antilog by using simple calculator

    ReplyDelete
    Replies
    1. I don't think you can yar. Scientific calculator or Log table is required.

      Delete
  48. Excellent and amazing!!!!!!!!!!!!!!!

    ReplyDelete
    Replies
    1. Thank you so much for appreciation. I hope you liked the video also.

      Delete
  49. What is the value of log(15.278)

    ReplyDelete
    Replies
    1. To do that, write it in scientific notation:

      log(15.278) = log(1.528 * 10^1) {Rounded off}
      = 1 + log(1.528) {properties of logs}

      Now, use log table according to what is mentioned in the article:
      1 + 0.1841
      = 1.1841 is the answer.
      Best of luck

      Delete
  50. dHANVAAD VEER JI....
    GOOD WORK BROTHER...
    APPRECIATE GHAT

    ReplyDelete
    Replies
    1. Koi chakkar ni bhaji. You're welcome

      Delete
  51. 196860÷2.303×8.314×298=logK

    ReplyDelete
    Replies
    1. It's kind of tricky question, but let's try. Please verify the answer, found by me, from somewhere else too. We can solve this, using 2 methods, either by calculating and then using logs, or only by using logs. I'm going to use the latter. Equation is:

      196860÷2.303×8.314×298=logK

      Take log on both sides & use logarithmic properties. {We've not taken antilog on both sides, because we don't know the properties of Antilogs}

      Log(196860) - Log 2.303 + Log 8.314 + Log 298 = Log(Logk)

      Solve Left hand side, by finding logs using the method shown in article.

      5.294 - 0.362 + 0.92 + 2.474 = Log (log k)
      8.326 = Log (log k)

      Now find anti-log of both sides,
      Al(8.326) = Log k

      Again find antilog on both sides,
      Al [Al(8.326)] = k

      Therefore first find Antilog of 8.326 & do that again on the result obtained to get the answer.
      k = Al[211836113.525]
      = 10^211836113 + Al(0.525)
      = 3.35 * 10 ^ 211836113

      which should be the approximate answer. It should be correct for only upto 4 digits of power, because of limitation of Log Tables & rounding off. I hope this helped.

      Delete
    2. Y do v have solve it using antilogs 2 times??

      Delete
  52. thank you it helped so much!...really thanks Mohak!

    ReplyDelete
    Replies
    1. You're welcome! I'm happy to help.

      Delete
  53. This comment has been removed by the author.

    ReplyDelete
  54. THANKS A LOT SIR. IT HAD HELPED ME IN SOLVING PROBLEMS OF CHEMISTRY

    sir i want to know what exactly exponential function and exponent mean .And what is log and why log

    ReplyDelete
    Replies
    1. Sure,
      Like a simple function F(x) can double the value of x as: F(x)=2x

      Logs, Antilogs & Exponential are also functions, but they don't have a formula(as far as I think), their value is different at different values of x. Check out the graphs of these(log (x), al(x) & e^x ; with x).

      The reason why these things have been made is that they're very useful in solving equations. For example the properties of logs can tell us the unknown power in the equation.
      (the hint of this, is in the video also)

      Then, the opposite of logs is also needed (Antilogs) to get the final answer.

      Exponential is also another function. It is related to the special case of Logs: Natural Log. Again it is also a very helpful function as its integral & differentiation remains the same. So it is also very useful in things like Laplace & Fourier Transforms. Wait! I don't think I've told, what Laplace & Fourier Transforms are!

      Anyways, just know that exponential function helps to solve integration & differential equations easily.

      This answer only explains a few uses of Logs, Antilogs & Exponential Functions, but they're used in a lot of applications, you will be using them a lot, in future also, I think.
      You can always use wiki for details ;)

      I hope you got the answer, you needed.

      Delete
  55. was really a good guidance.....good job!

    ReplyDelete
  56. sir can you please tell me how to find ans of root of 0.7278 using log and antilog....

    ReplyDelete
    Replies
    1. Sure,

      Log of root means Log [0.7278 ^ (1/2)].

      Now use logarithms' property, to get multiplication out of the bracket.
      It would now mean: 1/2 * log(0.7278)

      Now make use of log table to calculate value & then multiply by 1/2 to get the answer: -0.069 approximately.

      I hope you got it. If you didn't make sure you ask again.

      Delete
  57. awesome explanation dude!!!....and yes, this article is very important for students of class 11 and 12, especially in india. thank you very much!!

    ReplyDelete
  58. Thank u so much ... it's really helpful it was very easy to understand the way u taught 🌟🌟🌟🌟🌟

    ReplyDelete
    Replies
    1. You are welcome. I'm really happy, you liked it.

      Delete
  59. i owe my 12th chemistry grades to you sir ... hats off :)

    ReplyDelete
    Replies
    1. That's such a nice thing to tell!! Thank you for appreciation

      Delete
  60. squar root of 132 by log tabal

    ReplyDelete
    Replies
    1. Sure,

      Let x= 132 ^ (1/2)

      Take log on both sides & use properties of logs

      log x = 1/2 * log(132)

      Use log table & calculate,

      log x = 1.06 (approximately)

      Now, antilog both sides,

      x = al(1.06)
      =11.48, which is the approximate answer :-)

      Delete
  61. antilog not given properly

    ReplyDelete
  62. *phew* Chemistry exam ! Here I come ! :')

    ReplyDelete
  63. Chemistry Exam! Understood everything! Thanks for good Explanation!☺
    I have a doubt that according to explanation in taking antilog of something (eg. -8.654 ) we take a whole no and add and subtract to it. ( in this case we take 9). But if we have to take antilog of -4.4 the no. Used is 1 in one of my academic book. So please try to clear this Doubt.
    Thanku!

    ReplyDelete
  64. Why the log(.9876)= '-'0.0054? From where the negative sign comes???

    ReplyDelete
  65. How to remove logs of negative numbers?

    ReplyDelete
  66. how to calculate 3^1/3.146

    ReplyDelete
  67. Hi mohak
    It helped me a lot
    Explained better than chem teachers
    Can u pl tell me how to find antilog of 10^6

    ReplyDelete
  68. Hey I needed help with log base10 of a negative number. Like log (-3.8)

    ReplyDelete
  69. Hey I needed help with log base10 of a negative number. Like log (-3.8)

    ReplyDelete
  70. Well..explained.... Truly understood at a glance

    ReplyDelete
  71. How to know about characteristics in antilogarithms
    So that to put decimal

    ReplyDelete
  72. How to know about characteristics in antilogarithms
    So that to put decimal

    ReplyDelete
  73. Can't get the antilog will you please explain it clearly

    ReplyDelete
  74. sir, i m doing b.com in 2nd year from soldu. i want to know that log and antilog table will be given by the board in my exam or not?

    ReplyDelete
  75. Awesome explainations!
    But. How to find log 0.5?Plzzz help for my tmrws exam. .

    ReplyDelete
    Replies
    1. log0.5
      =log(5/10)
      =log5-log10
      =0.6990-1
      =-0.301

      Delete
  76. thank you sir!

    ReplyDelete
  77. How to find the antilog of 0.75, if the base of the log is 2401.
    Please answer, it's urgent for my upcoming test.

    ReplyDelete
  78. Well!! It really helped me alot. Still I have one doubt that what if bar comes on the number. So, how to find the log and antilog?

    ReplyDelete
  79. very valuable

    ReplyDelete
  80. How to find the antilog of 112.57

    ReplyDelete
  81. How to find log of single digit number...pls explain in deep..ex for log4 is 6020 but in log table its 6021? How to conclude this anu trick?

    ReplyDelete
  82. how to find log having more than 5 digits like 298.567

    ReplyDelete
  83. Such a clear explanations you have given. Thanks a lot.

    ReplyDelete
  84. thanks a lot . you explained well. but i have some doubt with AL(0.013). please help me.

    ReplyDelete
  85. what is the log(0.0064) by using table and calulator

    ReplyDelete
  86. Everything is just copied from here
    This guy knows nothing
    http://www.kareducation.in/2014/09/logarithms-antilogarithms-how-to-find-logs-antilogs-using-log-table.html?m=1

    ReplyDelete
  87. Please give me a link to download pdf file of logs (with up and down values of mean difference for first 19 numbers) and antilogs.

    ReplyDelete
  88. Finding logarithm is very clear and precise well i must admit that anti logarithm is a bit tough and unclear for a student not familiar with these

    ReplyDelete
  89. Thank you very much for saving my life!!! :)

    ReplyDelete
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