# Standard way to represent logarithms

What is the best/most correct way to represent the logarithm of a number? Example: $$-3 \log⁡2+5 \log⁡175+2 \log⁡7429+3 \log⁡34749$$

1. Just leave it the way it was calculated $$-3 \log⁡2+5 \log⁡175+2 \log⁡7429+3 \log⁡34749$$
2. As a single $$\log$$ $$\log \biggl(\frac{380082516906650443140753544921875}{8}\biggl)$$
3. Two logarithms for the positive and the negative part (in case they both exist. Otherwise use the above) $$\log 380082516906650443140753544921875 - \log 8$$
4. As a sum of logarithms of prime numbers $$-3 \log2 + 15 \log3 + 10 \log5 + 5 \log7 + 3 \log11 + 3 \log13 + 2 \log17 + 2 \log19 + 2 \log23$$
5. As a sum of logarithms with different coefficients ($$a\log b$$ means $$b$$ is the product of the primes numbers with exponent $$a$$ in the prime factorization) $$-3 \log2 + 15 \log3 + 10 \log5 + 5 \log7 + 3 \log143 + 2 \log7429$$

Is any of these the best way? Does it matter?

EDIT: I got curious because this number is really big. Obviously if the number is smaller the 2nd or 3rd options are fine

• I personally prefer $$\log \biggl(\frac{380082516906650443140753544921875}{8}\biggl)$$ but a nice feature of mathematics is that you can use any of the alternatives you mentioned and they all mean the same thing. – Maximilian Janisch Aug 15 at 10:05

Sometimes, you want a sense of how large this number is (such as if it appears as some measurement), in which case something like $$72.94$$ (I used WolframAlpha to compute this, assuming $$\log$$ is the natural log; if you want the base-10 log it's more like $$31.68$$) would actually be most appropriate. Sometimes, you just want that the number exists, in which case you can leave it at 1.