1. ## Problem with logarithm

If only log(2) and log(3) are given, then how can I calculate log(20) and log(30) ?

(The base is 10)

Thanks.

2. I believe you need $\displaystyle \log{(5)}$ too, since $\displaystyle 20 = 2 \times 2 \times 5$ and $\displaystyle 30 = 2 \times 3 \times 5$.

3. Originally Posted by Bacterius
I believe you need $\displaystyle \log{(5)}$ too, since $\displaystyle 20 = 2 \times 2 \times 5$ and $\displaystyle 30 = 2 \times 3 \times 5$.
... if the base is $\displaystyle 10$ , then is $\displaystyle \log 5 = \log 10 - \log 2 = 1 - \log 2$ ... but more simple is $\displaystyle \log 20 = 1 + \log 2$ and $\displaystyle \log 30 = 1 + \log 3$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by chisigma
... if the base is $\displaystyle 10$ , then is $\displaystyle \log 5 = \log 10 - \log 2 = 1 - \log 2$ ... but more simple is $\displaystyle \log 20 = 1 + \log 2$ and $\displaystyle \log 30 = 1 + \log 3$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Thanks a lot... that helps !

5. Actually, you don't need log(5). log(30)= log(3(10))= log(3)+ log(10) and log(20)= log(2(10))= log(2)+ log(10)- and you know log(10).

(Which I notice now is exactly what chi sigma said!)