**Is there a way to find exact log and antilog of a number without using log tables and calculators?**

In short : No you can’t find the exact values.

But yes you can find approximate values. Thanks to a bit of calculus. If you know the log values of first few prime numbers, you can get a close enough answer. All you need to do is express the number in a desirable product form (composed of smaller numbers whose values you perhaps know). This will lead you to a close value (still not good enough).

For example given you know log5, and you want to know log (5291), u get log(5.291×10^3) = 3+ log(5.291). Since you know log5 , you can approximate log(5.291) by Taylor expanding the log function about 5.

For better accuracy you should include more number of terms in this series.

**There’s a cool trick for finding log!**

log(1+x)≈0.4343(x−x22)log(1+x)≈0.4343(x−x22)

≈0.2171(x)(2−x)≈0.2171(x)(2−x)

Where 0≤x<<10≤x<<1

**Now how to use it?**

**Example: find log(8.256)log(8.256)**

log(8+0.256)=log(8(1+0.032))=3log2+log(1+0.032)log(8+0.256)=log(8(1+0.032))=3log2+log(1+0.032)

≈3(0.301)+0.2171(0.032)(1.968)≈3(0.301)+0.2171(0.032)(1.968)

=0.903+0.0137=0.903+0.0137

=0.9167

**Important log and anti-log values to learn before going for exam**

log2=0.301

log3=0.477

log5=0.6987

log7=0.845

e=2.718

e^2=7.389

The above values are more than enough for Aipmt paper. Other values can be found out using these results.

For eg -log15=log3×5=log3 + log5

Most of the logs you can find out by remembering log2,log3,log5 and log7.

Use the properties: log(ab)=log(a) +log(b), so log(6)=log(3)+log(2)

also obviously, log(abab)=b××log(a) , so log(8)=log(2323)=3××log(2)

As for antilogs, the basic property is y=log(x) => x=eyeyOR 10y10y if you are using log10(x)log10(x). These methods will not obviously work every time, but they do most of the time.

As a culmination of these methods and approximation, let’s try to find log(893)

log(893)=log(127.57××7)≈≈log(7)+log(128)=log(7)+log(2727)=log(7)+7××log(2)=0.845+7××0.301=2.952 (actual value of log(893)=2.951)

**How can I quickly find log and antilog values during exams?**

ANTILOG: This is easiest of all

Example: . log(x) = y here log of x is y , similarly ANTILOG of ‘y’ is x

So our main task is to find value of ‘x’

log(x) =y Then 10^y = x (we got Value of x)

- Sums

Find ANTILOG (5) ans = 10^5

Find ANTILOG(2). Ans = 10^2=100

- FORMULA : ANTILOG(m) = 10^m

There’s a quick trick someone told me and it’s helping me till date.

for eg, log2(x)=10 Find value of x?

Here “2” is the base

Then, imagine pushing that 2 from below the log. This 2 will go beyond equal to sign and further pushes the 10 above it. As soon as you do this, remove the log.

Now the equation will look like, x=2^10

**How will I calculate antilog (4.33) in exam?**

The no.before point becomes power of ten.

Now .33 is left.

Log2=0.3010

Log3=0.4771

.33 lies near to log 2.

Approx. 2.3

So ans will b 2.3 *10^4

**How do I find log and antilog of negative numbers?**

Log of negative numbers is not defined.

For antilog of negative numbers write the number x as [x] + {x}.

[x] will give you the exponent to tha base 10 and antilog of {x} found using antilog table will give you coefficient.

For example

antilog (-0.699)

= antilog (-1+0.301)

= 2x 10^(-1)

(As log 2 = 0.301, antilog (0.301)= 2)

**LOGARITHM :**

- THERE IS NO FORMULA TO GET ANS WITHIN SECONDS ..ALL METHODS MENTIONED BELOW ARE USED TO CALCULATE LOG .MANY OF WHICH USED IN CALCULATORS AND OTHER IS THE LOG TABLE. APPROXIMATIONS CAN BE MADE USING TAYLOR SERIES( TAKES LOT OF TIME). Cannot be used for calculating log in ionic and chemical equilibrium chapters. U need to remember some values such as log2 log3 etc. But in exam they provide values which u cannot predict {end} ..the content below is extra if u need.

Logarithms are easy to compute in some cases, such as log10(1000) = 3 or simple log values can be found using properties of log. In general, logarithms can be calculated using **Power series** or the **arithmetic- geometric mean** or be retrieved from a precalculated . that provides a fixed precision.

**NEWTON’S METHOD**, an iterative method to solve equations approximately, can also be used to calculate the logarithm, because its inverse function, the exponential function, can be computed efficiently.

Using look-up tables, **CORDIC**-like methods can be used to compute logarithms if the only available operations are addition and bit shifts

Thanks a lot for valuable knowledge

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