Question

Please help me

How can we calculate log and antilog of a number WITHOUT using log tables ?
The exam I am preparing for does not allow using log tables.

Answer 1

the log tables were created literally through approximating with square roots until they got closer and closer to the actual value. It would be extremely difficult for you to do so unless you are allowed a basic calculator. In which case, you can use guess and check to 3 decimals. Although through experience, exams will not make you do this. You should probably expect questions where you can reason what the log is. like log 1000 = 3....
or log (base 2) of 30 = log30/log2 and you can leave it there.

Another way of solving such problems is when you memorize basic logs of log2,log3,log4,log5. You can cleverly manipulate their logs to formats using these and then just simplify.

Answer 2

you can use a polynomial to approximate it for a given interval and then determine the level of accuracy you want without a table

Answer 3

Thanks a lot Puzzhang
It will surely help me

But the exam involves large calculations and also down't allow us the use of a calulator
and moreover time is a big factor in that exam too

Answer 4

@ amistre Can u explain it please

Answer 5

well, it doesnt make it easier perse; but it does make it doable by hand ...

Answer 6

derivative: ln(1+x) = 1/(1+x)


1-x+x^2 -x^3+x^4 ...
----------------
1+x ) 1
(1+x)
------
-x
(-x-x^2)
--------
x^2

\[\int 1-x+x^2 -x^3+x^4 ...dx\]
\[x-x^2/2+x^3/3 -x^4/4+x^5/5 ...\to\ \sum_{n=1}^{inf}\frac{(-1)^{n+1}x^n}{n}=ln(1+x)\]

Answer 7

Oh u mean I would have to use expansions

Answer 8

now the issue is how accurate you need your results to be ...

we can determine how many of these terms we need to get to a certain level of accuracy

Answer 9

But what about antilog

Answer 10

the antilog is a polynomial function of for "e"

Answer 11

Yaaaa
Thanks a lot amistre it really helped me

Answer 12

youre welcome