Bases, Exponents, and Digits

Jan 2016
19
1
US
Question: What's the smallest positive value of k such that the base nine representation of $\left ( 27_{ten} \right )^{k}$ has exactly 9999 (base ten) digits?

Work thus far: I know that in base ten to find the number of digits you just take the common log and round up, and I'm guessing that here you just take the log base nine of the number. I tried coming up with equations like $k\log_{9}27 = 9999$ meaning k = 6666 but I'm still kind of stuck and confused. Ans key says k is 19996/3, which is close to my answer.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey cake27.

If you have a number (call it N) and you want to find the number of digits in some fixed base then log_b(N) will tell you that. If it has a fractional component then you have to basically add another digit after you remove the non-integer part. First I'm going to find the number of base 9 digits and I encourage you to to the same but for base 10.

So basically you finding log_9[27^k] = klog_9[27] = 9999.

Now since 27 = 3^3 this means

k
= 9999 / log_9[27]
= 9999 / log_9[9*3]
= 9999 / [1 + log_9[3]]
= 9999 / [1 + 0.5] since 9^(0.5) = 3
= 9999 / [1.5].

See if you can do the same thing but for base 10 instead of base 9.

You should find the answer should be less than base 9 since the index is higher.
 
Jan 2016
19
1
US
Wait, but 1 + log_10[2.7] doesn't simplify to an exact form?
 
Dec 2012
1,145
502
Athens, OH, USA
Cake,
Here is some discussion about integers base b and a "derivation" of the text answer. But it turns out that there is no integer k satisfying the problem!

 
Last edited:
Jan 2016
19
1
US
Ok, thanks! Makes complete sense now. (Rofl)