Question

The number of digits in ${20^{301}}$ ( given , ${\log _{10}}2 = 0.3010$) is
A.$602$
B.$301$
C.$392$
D.$391$

Answer 1

Hint : In the question we have given the value of logarithm $2$ , therefore we must use the concept of logarithm to solve it . We will use the product law for logarithm and exponent law of logarithm to solve it . The number of digits will always be one greater than the characteristic. Always use the given value of logarithm in the question . Remember the laws for logarithm and exponent . In a logarithm number the value left of the decimal place is called characteristic and the value right of the decimal is called mantissa .

Complete step-by-step answer :
Given : ${20^{301}}$ , ${\log _{10}}2 = 0.3010$
Let $x = {20^{301}}$
Taking $\log$ on both sides we get , here we have used base $10$ .
$\log x = \log {20^{301}}$
Now using the laws of exponent for logarithm $\log {m^n} = n\log m$ on LHS we get ,
$\log x = 301\log 20$
On simplifying we get ,
$\log x = 301\log \left( {2 \times 10} \right)$
Now using the product law for logarithm $\log m \times n = \log m + \log n$ we get ,
$\log x = 301\left[ {\log 2 + \log 10} \right]$ ,
Now using the given value of ${\log _{10}}2 = 0.3010$ and ${\log _{10}}10 = 1$ in the above equation , we get
$\log x = 301\left[ {0.3010 + 1} \right]$
On solving we get ,
$\log x = 301\left[ {1.3010} \right]$
On solving again we get ,
$\log x = {\text{391}}{\text{.601}}$
The characteristic part has value $= 391$
According to hint , the required value will be one greater than , therefore $= 391 + 1$ ,
$= 392$
Therefore , option ( 2 ) is the correct answer for the given question .
So, the correct answer is “Option 2”.

Note : The question can be calculated using integration method which is as follow :
Number of digits = integrated part of $\left( {301\log 20} \right) + 1$
On solving we get ,
integrated part of $301\log \left( {2 \times 10} \right) + 1$
integrated part of $301\left[ {\log 2 + \log 10} \right] + 1$
integrated part of $301\left[ {1.3010} \right] + 1$
integrated part of $\left( {{\text{391}}{\text{.601}}} \right){\text{ + 1}}$
$= 391 + 1$
$= 392$
Hence Proved .