The question is as follows:
The expansion of $(\frac{2}{3})^{30}$ begins with $0.000…$ . How many zeros are there between the decimal point and the first nonzero digit? In the expansion of $(\frac{2}{3})^{1000000}$ how many zeros are there between the decimal point and the first nonzero digit?
I used my calculator to solve for the value of $(\frac{2}{3})^{30}$ and I got that there are $5$ numbers that are in between the decimal point and the first nonzero digit. I thought that maybe taking $\log((\frac{2}{3})^{1000000})$ for which I would get $-176091.2591$. I know that if you have $10^n$, then number of digits that that value will have is going to be $n + 1$. I was thinking that maybe I can use the same principle in answering this question because $10^{-176091.2591} = (\frac{2}{3})^{1000000}$, so there would be $176092$ digits. If I am wrong, can someone please correct me or help me? Any help will be greatly appreciated!
Best Answer
The numbers with $n$ zeros between the decimal point and the first nonzero lie in the interval $[10^{-(n+1)}, 10^{-n})$. So given a number $x$, the number of zeros is $\lceil - \log_{10} x \rceil - 1$.
Indeed $\log_{10}((2/3)^{30})=-5.28$ which gives you the $5$ zeros. Similarly, $\log_{10}((2/3)^{10^6}) = - 176091.3$ which gives you $176091$ zeros.