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


*

*$9^1=9$

*$9^2=81$

*$9^3=729$

*$9^4=6561$
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
