Number of digits in $8^n$ in base $6$ Let $d(n)$ be the number of digits when $8^n$ is written in base $6$. Find the closed form expression for $d(n)$.I tested for few numbers:
$$8^1=12$$
$$8^2=144$$
$$8^3=2212$$
So i felt that $d(n)=n$. But to formally prove it, i tried with induction.
Let $d(n)=n$, then we have
$$8^n=a_0a_1a_2....a_{n-1}, \: a_i \in \left\{0,1,2,3,4,5\right\}, \:a_0 \ne 0$$
Now we have
$$8^{n+1}=8^n \times 12$$
I am stuck here?
 A: the number of digits of an integer $a$ in base $b$ is $\lfloor\log_b(a)\rfloor+1$ (because it is constant on the interval $[b^k,b^{k+1})$ and jumps by $1$ at values of the form $b^k$).
Hence we want $\lfloor\log_6(8^n)\rfloor+1= \lfloor n\log_6(8)\rfloor +1 = \lfloor n \frac{\ln(8)}{\ln{6}}\rfloor+1$
we have $\frac{\ln{8}}{\ln{6}}\approx 1.16055842$
A: Hint: $8^n = 6^{(\log_6{8})n} = 6^{1.1605\ldots n}.$
So for example, $8^{100} = 6^{116.05...},\ $ so this number has $117$ digits base $6.$
A: $d(n) = n+1$ only works for small values of $n$.  Note the leading digit is creeping up eventually you will have an $8^k = 5xxxx...x$ with $k+1$ digits but $8^{k+1} = 1yxxxx....x$ with $k + 3$ digits.
Indeed $8^6 = 5341344$ with $7$ digits and $8^7 = 112541012$ with $9$ digits.
=====
You can think backwards.  If $8^n$ has $K$ digits then $8^n \ge {1\underbrace{000.....0}_K}_{base\ 6} =6^{K-1}$ and $8^n \le {\underbrace{5555.....5}_K}_{base\ 6}=6^{K}-1$ so $6^{K-1} \le 8^n < 6^K$.
So solve for $K$

 $\log_6 6^{K-1} \le \log_68^n < \log_6 6^K$

$K-1 \le n\cdot \log_6 8 < K$

$K-1 = \lfloor n\cdot \log_6 8 \rfloor$

  $K = \lfloor n\cdot \log_6 8 \rfloor

