Question

How do you calculate the antilog of 2?

Answer 1

Hint: To calculate the antilog of 2, we are going to use the following relation in log which states that: $\log b=a$ and here we are taking the base of the log as 10. Then antilog of $a$ is b. Now, we know the logarithm property that if $\log b=a$ then $b={{10}^{a}}$ and this value of b is the antilog of a. So, to find the antilog of 2 we are going to substitute 2 in place of $a$ in the above formula.

Complete step by step answer:
In the above problem, we are asked to calculate the value of:
antilog of 2
We know the logarithm property that:
$\log b=a$
In the above equation, we are taking the base of the log as 10 because in general the base of the log is 10. Then the antilog of $a$ is equal to ${{10}^{a}}$ which is equal to b.
Now, to find the antilog of 2 we are going to put $a$ as 2 in ${{10}^{a}}$ and we get,
${{10}^{2}}$

From the above, we got the antilog of 2 is ${{10}^{2}}$.

Note: The value of antilog of 2 that we are getting is correct or not, we can check it in the following way:
We have calculated the value of antilog of 2 as ${{10}^{2}}$. Now, equating the value of antilog of 2 to ${{10}^{2}}$ we get,
$anti\log \left( 2 \right)={{10}^{2}}$
Taking log on both the sides we get,
$2=\log {{10}^{2}}$
As the base of the log is 10 and we know the property of the logarithm that:
$\begin{align}
  & {{\log }_{a}}{{a}^{b}}=b{{\log }_{a}}a \\
 & \Rightarrow {{\log }_{a}}{{a}^{b}}=b \\
\end{align}$
So, using the above property of logarithm in $2=\log {{10}^{2}}$ we get,
$2=2$
Now, L.H.S = R.H.S so the value of antilog (2) that we have calculated above is correct.