0

While determining number of digits in $3^{43}$ I have solved it by taking $\log_{10}$ both sides: \begin{align} x = 3^{43} \\\ \log_{10} x = \log_{10} 3^{43} \\\ \log_{10} x = 43\log_{10} 3 \\\ \log_{10} x = 43 * 0.4771213 \\\ \log_{10} x = 20.5162159 \\\ \end{align} which means it has 21 digits. However, I was wondering what if instead of taking $\log_{10}$ both sides I take $\log_3$ both sides because then $\log_3 3$ will be equal to 1: \begin{align} x = 3^{43} \\\ \log_3 x = \log_3 3^{43} \\\ \log_3 x = 43 \\\ \end{align} So can't we determine number of digits in $3^{43}$ from above equation? What difference will it make in this case?

19aksh
  • 12,768
Anurag Gupta
  • 79
  • 1
    Because what you did has nothing to do with the number of digits. Since you use base 10 numbers, you need to involve 10 in your calculations. Where is that coming in into $\log_3 x$? – Andrei May 01 '20 at 03:43
  • I got your point. But that means that when we use 'e' in log i.e. $\log_e$ we use a number system whose base is 'e'? – Anurag Gupta May 01 '20 at 03:51
  • No. What I meant in this case is that whatever number you get, you can write it as $3^{43}=a.bcde...\cdot 10^x$. Here $a,b,...$ are from $0$ to $9$, except $a$ which has to be from $1$ to $9$. $x$ is an integer, and will be the number of digits. That's why you need to take logarithm base 10, so the expression will be $$\log_{10}3^{43}=\log_{10}a.bcd...+x$$Here the remaining log is greater than $0$ (since $a\ge 1$) and less than $1$ – Andrei May 01 '20 at 04:01
  • I've added my comment as an answer – Andrei May 01 '20 at 04:02

2 Answers2

1

That would give the number of digits if you were writing it in base $3$.

(Although you would actually have to use $44$ digits. You can see this by checking what happens in base $10$, where $10^2$ is the first number that requires $3$ digits to write. To be 100% accurate you could express this mathematically as "The number of digits required to write $N$ in base $b$ is the smallest integer strictly greater than $\log_b N$".)

JonathanZ
  • 10,615
  • So that means that when we use 'e' in log i.e. $\log_e$ we use a number system whose base is 'e'? – Anurag Gupta May 01 '20 at 03:53
  • 1
    Well, logarithm's main use isn't figuring out how many digits you need to use when writing numbers. But if you want to have some weird fun, go search for and read about "non-integral number bases". – JonathanZ May 01 '20 at 03:56
0

You can write your number as $$3^{43}=a.bcde...\cdot 10^x$$ Here $a,b,...$ are from $0$ to $9$, except $a$ which has to be from $1$ to $9$. $x$ is an integer, and will be the number of digits. That's why you need to take logarithm base 10, so the expression will be $$\log_{10}3^{43}=\log_{10}a.bcd...+x$$Here the remaining log is greater than $0$ (since $a\ge 1$) and less than $1$

Andrei
  • 37,370