Question

Find the number of digits in integral part of ${{60}^{12}}+{{60}^{-12}}-{{60}^{-15}}$. (given $\log 2=0.3030,\log 3=0.4771$)
A. 20
B. 21
C. 22
D. 24

Answer 1

Hint: We need to find the terms responsible for the part before the decimals. We then try to form it as a power of 10. We use different laws of logarithm using base 10. At the end we find a number of terms in the expansion of 10 with its unique power number which gives the solution.

Complete step-by-step answer:
There are binary operations addition and subtraction of three different terms.
We will try to find the term responsible for the integral part of ${{60}^{12}}+{{60}^{-12}}-{{60}^{-15}}$.
Let’s take the last two terms. Those have negative integers on their power.
We need to convert the power into positive terms using the law of exponents.
We know that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ where $a\ne 0$.
So, both ${{60}^{-12}}$ and ${{60}^{-15}}$ can be expressed like
${{60}^{-12}}=\dfrac{1}{{{60}^{12}}}$ and ${{60}^{-15}}=\dfrac{1}{{{60}^{15}}}$.
As the terms ${{60}^{12}}$ and ${{60}^{15}}$ are greater than 1. So, $\dfrac{1}{{{60}^{12}}}$ and $\dfrac{1}{{{60}^{15}}}$ are less than 1.
So, in the whole part of ${{60}^{12}}+{{60}^{-12}}-{{60}^{-15}}$, only ${{60}^{12}}$ is responsible for the terms before decimal as rest is responsible for the part after decimal.
We need to find out the number of terms in the expansion of ${{60}^{12}}$.
Let’s assume $P={{60}^{12}}$. We take logarithm with base 10 on both sides.
So, $\log P=\log \left( {{60}^{12}} \right)$.
Now we know $\log \left( {{a}^{b}} \right)=b\log a$, $\log \left( xy \right)=\log x+\log y$, ${{\log }_{a}}a=1$.
Applying these rules, we get
$\begin{align}
  & \log P=\log \left( {{60}^{12}} \right) \\
 & \Rightarrow \log P=12\log 60 \\
 & \Rightarrow \log P=12\log \left( 10\times 2\times 3 \right) \\
 & \Rightarrow \log P=12\left[ \log \left( 10 \right)+\log \left( 2 \right)+\log \left( 3 \right) \right] \\
\end{align}$
Now the values are given $\log 2=0.3030,\log 3=0.4771$ and also ${{\log }_{10}}10=1$.
We put the values
$\begin{align}
  & \log P=12\left[ 1+0.3030+0.4771 \right] \\
 & \Rightarrow \log P=12\left[ 1.7801 \right]=21.3612 \\
\end{align}$
Now, as the base of the logarithm is 10, we try to reverse it.
We know that ${{\log }_{z}}b=x\Rightarrow {{z}^{x}}=b$.
So, $\log P=21.3612\Rightarrow P={{10}^{21.3612}}$
Now any power of 10 indicates the number of zeros in that number. Decimal numbers won’t affect that number as for that we need integers.
So, in P the number of integers before decimal is equal to the number of digits in ${{10}^{21.3612}}$.
Now in finding the number of digits we only need ${{10}^{21}}$.
So, we got 21 times 0’s and started number 1.
In total there will be 22 numbers.
So, the number of digits in integral part of ${{60}^{12}}+{{60}^{-12}}-{{60}^{-15}}$ is 22.

So, the correct answer is “Option C”.

Note: We need to understand that the real value of that power of 10 is not necessary for the solution. The power of 10 defines the number of terms there will be in the expansion and that’s the most important thing. We need not to worry about the decimal part of the power as it doesn’t change the number.