Find the smallest natural number n such that the decimal expansion of $9^n$ has $< n$ digits?

We can see that,

1. $9^1=9$
2. $9^2=81$
3. $9^3=729$
4. $9^4=6561$ if there exists such n then what is the prove. Thank you.

The base 10 log shows us how many digits there are in the decimal expansion. Notice how $log_{10}100=2$

$log_{10}1000=3$

$log_{10}500=2.7$

The number of digits in $x$ is given by the formula:

$\text{digits}=\lfloor{log_{10}x}\rfloor+1$

Where $\lfloor{z}\rfloor$ is the number $z$ rounded down to the nearest integer.

So for your question you can use this formula:

$\lfloor{log_{10}9^n}\rfloor+1 < n$

Use laws of logs to simplify:

$\lfloor{nlog_{10}9}\rfloor < n-1$

From the definition of the floor function clearly

$\lfloor{nlog_{10}9}\rfloor \leq nlog_{10}9 < n-1$

Therefore we can get a lower bound for $n$ by solving

$nlog_{10}9 < n-1$

$1 < n(1-log_{10}9)$

$21.85 < n$

Therefore you only have to search for $n \geq 22$

You can see that $9^{22}=9.85\times 10^{20}$ which clearly has 21 digits