To determine number of digits in $3^{43}$

Question

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?

Answer 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 $\log_b N$".)

Answer 2

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 $1$