$$\begin{align} \lg&=\log_{10}\\ n&:=\text{number of the digits of}~7^{35}\\ 35\lg 7&<n\in\mathbb N\\ 35\lg 7&<n \underbrace{\leq 35\lg 7 +1}_{\text{what i want to prove} } \end{align}$$
At least I know that $~ \lg7 <1 ~$ is true, and I've been got stuck.
Can anyone give me some hint(s)?
On
In general, we could prove that if a positive integer $M$ has $n$ digits, then $\,\log_{10}M<n\leqslant\log_{10}M+1\,.$
Proof :
The smallest positive integer number which has $n$ digits is $10^{n-1}$, whereas the greatest one is $10^n-1$.
Consequently, since $M$ has $n$ digits, it follows that
$10^{n-1}\leqslant M\leqslant10^n-1<10^n\;\,,\;\;\,$ hence ,
$\log_{10}10^{n-1}\leqslant\log_{10}M<\log_{10}10^n\;\;,$
$n-1\leqslant\log_{10}M<n\;\;\,,\;\;\;$ which is equivalent to
$\log_{10}M<n\leqslant\log_{10}M+1\,.\\[10pt]$
In particular, if $M=7^{35}$, we get that
$\log_{10}7^{35}<n\leqslant\log_{10}7^{35}+1\;\;\,,\;\;\;$ that is ,
$35\log_{10}7<n\leqslant35\log_{10}7+1\;.$
If you calculate $7^7$ by hand, you see that it is equal to $8.23\times 10^5$. This is strictly between $8\times 10^5$ and $10^6$.
Since $7^{35}=\left(7^7\right)^5$, we see that $$7^{35}<10^{30}$$ We also have $$7^{35}>8^5\times 10^{25}$$ Since $8^5=2^{15}=32\cdot 1024>3\times 10^4$, we have that $7^{35}>10^{29}$. Hence it has $30$ digits.