Question

How do you calculate $anti\log \left( 0.6117+\text{ }-3 \right)$ ?

Answer 1

Hint: There is no provision in the conventional logarithmic tables to locate the antilogarithm of a negative number. Thus, we have to manipulate the given negative number whose antilogarithm is to be found to calculate the antilogarithm. Therefore, we shall first convert the decimal part of the given number as a positive number and then look for its antilogarithm value in the antilogarithm table.

Complete step by step solution:
Given to calculate antilogarithm of $\left( 0.6117+\text{ }-3 \right)$ which is equal to antilogarithm of $\left( -2.3883 \right)$.
 $\Rightarrow anti\log \left( 0.6117+\text{ }-3 \right)=anti\log \left( -2.3883 \right)$
There are two parts of the given negative number. The first part is before the decimal point and the second part is after the decimal point.
When the part before the decimal point is negative, it does not create any problem and it will not be modified as it will be assigned as power to the default base 10 of the logarithm and there is no problem with a negative power to base 10.
However, we will make changes to the part of the number after the decimal point as it has to be made positive.
The negative number $\left( -2.3883 \right)$ can also be written as $-3+0.6117$.
Now the base will be assigned a power of -3 and the remaining positive number 0.6117 will be located in the tables of antilogarithm in the row of 0.61 and in the column of 1.
We see that the antilogarithm value of 0.6117 is 4.0897.
Thus, the total antilogarithmic value becomes $4.0897\times {{10}^{-3}}$.
$\Rightarrow anti\log \left( 0.6117+\text{ }-3 \right)=4.0897\times {{10}^{-3}}$
$\Rightarrow anti\log \left( 0.6117+\text{ }-3 \right)=0.0040897$
Therefore, $anti\log \left( 0.6117+\text{ }-3 \right)$ is equal to 0.0040897.

Note: We understand antilogarithm functions as the inverse of the log functions. If ‘n’ is the logarithm of a given number ‘x’ with a given base, then ‘x’ is called the antilogarithm (or antilog) of ‘n’ to that base. That is, if ${{\log }_{a}}x=n$, then $n=anti\log \left( x \right)$.