Basic Logarithms problems

Question

Can someone tell me how many digits would be there in-

$(2.5)^{200}$ and $6^{50}? $

I'm utterly confused where to begin from. Any help would be appreciable.

Do you understand that the result will not be a whole number ? — Claude Leibovici, Jun 20 '16 at 07:32
Yes but that's where the characteristics and mantessa will come in play right? Please tell me how to proceed — Zlatan, Jun 20 '16 at 07:34
You will want to apply $ \ \log p^n \ = \ n \ \log p \ $ and the "number facts" that $ \ \log_{10} \ 2.5 \ \approx \ 0.4 \ $ and $ \ \log_{10} \ 6 \ \approx \ 0.78 \ $ . — colormegone, Jun 20 '16 at 07:36
Hint: Given $x^y$, find $z$ such that $x^y={10}^z$. Do you (a) see how to do that, and (b) see how $z$'s your answer? — John Forkosh, Jun 20 '16 at 07:37

score 2 · Answer 1 · answered Jun 20 '16 at 07:34

2

$6^{50}=10^{50/\log_{6}(10)}\implies6^{50}$ has $\lceil{50/\log_{6}(10)}\rceil=39$ digits.

answered Jun 20 '16 at 07:34

barak manos

score 1 · Answer 2 · answered Jun 20 '16 at 07:44

Another approach that shows the rule of ''characteristic'' and ''mantissa''

starting from: $\log_{10} 2.5 \approx 0.3979$

we have: $ \log_{10} (2.5)^{200} \approx 200 \cdot 0.3979 \approx 79.56=79+0.56 $

where $79$ is the characteristic and $0.56$ is the mantissa.

So: $ (2.5)^{200}\approx10^{79+0.56}=10^{79}\cdot 10^{0.56}=3.6 \cdot 10^{79} $

2 Answers2