3

$(1)$:: Calculation of no. of Digits in $2^{100}$ .$(2)$:: Calculation of no. of Digits in $3^{100}$.

If it is given that $\log_{10}(2)=0.3010$ and $\log_{10}(3) = 0.4771$

$\bf{My\; Try::}$ I have seen in book and it is given as ::

$(1)$ no. of Digit in $\displaystyle 2^{100}$ is equal to $\displaystyle \lfloor \log_{10}(2)^{100}\rfloor +1\;,$ where $\lfloor x \rfloor = $ floor function of $x$.

$(2)$ no. of Digit in $\displaystyle 3^{100}$ is equal to $\displaystyle \lfloor \log_{10}(3)^{100}\rfloor +1\;,$ where $\lfloor x \rfloor = $ floor function of $x$.

But I did not Understand How can we prove ..

no. of Digit in $(x)^y$ is equal to $\lfloor \log_{10}(x)^y\rfloor +1.$

plz explain me in Detail.

Thanks

juantheron
  • 53,015

2 Answers2

6

The integer $x^y$ will lie between two successive powers of $10$ (or successive powers of any other integer greater than $1$, for that matter), so you can find a unique integer $n$ such that $$10^n\leq x^y<10^{n+1}$$ This means that, in fact, $$10^n\leq x^y\leq 10^{n+1}-1$$ The number on the left has $n+1$ digits, and the number on the right has $n+1$ digits, so the number in the middle must also have $n+1$ digits.

What is $n$? Take logs base $10$ in the first displayed inequalities to see that $$n\leq \log_{10}x^y< n+1$$ This is exactly the definition of $$n=\lfloor\log_{10}x^y\rfloor$$ so we immediately have the desired result that the number of digits is $$n +1=\boxed{\lfloor\log_{10}x^y\rfloor+1}$$

MPW
  • 43,638
0

Use the fact that $\log$ is power required by $10$ to approach that number.

If the number is of $2$ digits the value is $\ge1,<2$. because $10^1=10,10^2=100$

evil999man
  • 6,018