Antilog of -1.6132 using anti-log tables

Question

I have different answers for antilog(-1.6132) using antilog tables and the calculator.

Using this site as reference https://byjus.com/maths/antilog-table/.

Mantissa: 6132
Characteristic: -1

Corresponding value against mantissa (using antilog tables):

.61[col:3] = 4102
.61[mean.diff.col:2] = 2

Required digits: 4102+2 = 4104

As characteristic = -1, antilog(-1.6132) = 0.4104

But, using a calculator, antilog(-1.6132) = 0.024366884

Obviously, I am missing something while looking up the antilog table. Any help is appreciated.

Antilog table's relevant part as image:

I am trying to exactly follow the steps:

Procedure to Find the Antilog of a Number

Method 1: Using an Antilog Table

Consider a number, 2.6452

Step 1: Separate the characteristic part and the mantissa part. From the given example, the characteristic part is 2, and the mantissa part is 6452.

Step 2: To find a corresponding value of the mantissa part use the antilog table. Using the antilog table, find the corresponding value. Now, find the row number that starts with .64, then the column for 5. Now, you get the corresponding value as 4416.

Step 3: From mean difference columns find the value. Again use the same row number .64 and find the value for column 2. Now, the value corresponding to this is 2.

Step 4: Add the values obtained in step 2 and 3, we get 4416 + 2 = 4418.

Step 5: Now insert the decimal point. The decimal point always goes the designated place. For this, you have to add 1 to the characteristic value. Now you get 3. Then add the decimal point after 3 digits, we get 441.8

So the antilog value of 2.6452 is 441.8.

Answer 1

When dealing with negative powers, you have to be careful.

It looks like you tried to compute $10^{-1.6132}$ by starting with $10^{0.6132} \approx 4.104$ (from a log table) and then shifting the decimal. However, shifting the decimal has the effect of subtracting an integer from the power: we have $0.4104 \approx 10^{0.6132 - 1} = 10^{-0.3868}$.

Instead, to compute $10^{-1.6132}$, you should start with $10^{-1.6132 + 2} = 10^{0.3868}$, which the method you're using will tell you is approximately $2.437$. Then, shifting the decimal two positions, we conclude $10^{-1.6132} \approx 0.02437$.

When we start with a negative number like $-1.6132$, and add an integer to it to put in the range $[0,1)$, we end up with a positive number with very different digits! Watch out for that.

---

In other words: just like we write $2.6452$ as $2 + 0.6452$ to see that it has a characteristic part of $2$ and a mantissa part of $6452$, we should write $-1.6132$ as $-2 + 0.3868$ to see that it has a characteristic part of $-2$ and a mantissa part of $3868$. The mantissa part should always come from a positive quantity between $0$ and $1$.

Answer 2

For those of us who actually used log-tables,

rather than $\text{antilog}(-1.6132)$, which has a negative characteristic and a negative mantissa when we prefer to work with a non-negative mantissa in $[0,1)$,

we would instead write $\text{antilog}(\overline{2}.3868) $ by making the characteristic more negative so the mantissa is in $[0,1)$, using the overline on the characteristic to show this. We then have $$\text{antilog}(-1.6132) = \text{antilog}(\overline{2}.3868) = \text{antilog}(0.3868) \times 10^{-2}.$$

For $\text{antilog}(0.3868)$, you look in the .38 row of your table and the 6 column to see 2432 and then at the 8 mean differences column to see 4, adding these together to get $2432+4=2436$ representing a value in ${[10^0,10^1)}=[1,10)$,

so you get $2.436\times 10^{-2}=0.02436$ as the result.

Answer 3

The proper approach is always to render the logarithm in the form

(Mantissa) + (Characteristic)

where

• (Mantissa) is between 0 and 1

• Characteristic is an integer

The first constraint means you can't carry a negative sign in the mantissa. When you have a negative number, you have to convert the decimal part to a positive number by taking the base-10 complement. Here, $6132$ with a negative sign becomes $10^4-6132=3168$, so properly the mantissa is $0.3168$. Clearly the characteristuc then has to be $-2$, as $-1.6132$ is two units kess than $0 3168$. So the antilogarithm is whatevervthe table says forca mantissa of $0.3168$ multiplied bt $10^{-2}$.

Answer 4

That site doesn't really define "mantissa", it just gives an example and expects you to infer what it means. A better term would be "floor". If we define "floor of x" as being the largest integer less than or equal to x, and "remainder" as the x minus the floor of x, then replacing "characteristic" with "floor" and "mantissa" with "remainder" gives an algorithm that gets the same answer for positive numbers, but, unlike the original, gets the correct answer for negative numbers as well. The floor of -1.6132 is -2, and the remainder is 0.3868. Plugging those numbers in gets the correct answer of 0.024366884 . The problem is that the floor is rounding down, while the characteristic is truncating, and when you truncate a negative number, you're actually rounding up.

The decimal point always goes the designated place.

That is a completely useless statement. "You should put the decimal point where the decimal point goes" is a tautology.

Characteristic Part – The whole part is called the characteristic part. If the characteristic of logarithm of any number greater than one is positive and is one less than the number of digits in the left side of the decimal point.

That's just complete gibberish. This may be trying to say that the case of negative numbers has to be treated differently, but the only way I'd read it that way is if I already know negative numbers have to be treated differently. This seems to be a non-English grammar applied to English vocabulary. Perhaps to people raised with the same mish-mash of languages this is informative, but you should seriously consider trying to find a source of Math explanations with more standard grammar, especially if you plan on using your Math in Western contexts.