| p | 2p | 3q | 6q | |||||
a) | b) | c) | d) | |||||
| 6q | 3q | 2p | p |
| 1 | ||
log725 = | ||
| log257 |
| 1 | ||
log257 = | log57 | |
| 2 |
| 3q | ||
po podstawieniu: = | ... odpowiedź: c) | |
| 2p |
| 6q | ||
sorry .... oczywiście: = | ... odpowiedź: d) | |
| p |
Mi wyszło tak jak Tobie na początku 3q/2p
| 1 | |||||||
* 3q = 2p * 3q = | |||||||
|
| 1 | 6q | |||
=3log7(3)*2*log7(5)=6*log7(3)* | = | |||
| log5(7) | p |
| 1 | 2 | 6q | |||||||||
*3q= | *3q= | ||||||||||
| p | p |
| p | 2 | 6q | ||||
( | )−1*3q= | *3q= | ||||
| 2 | p | p |