| 3(1−m) | ||
niech m = log217. Wykaż że log727 = | ||
| m |
| 3(1−m) | ||
czy mogę po prostu podstawić m do log727 = | i udowodnić że L=P? | |
| m |
| 1 | |
=log7 21 = log7( 7*3) = log7 7 + log7 3 = 1 + log7 3 | |
| m |
| 1 | ||
log7 3 = | − 1 | |
| m |
| 1 | ||
log7 27 = log7 33 = 3 log7 3 = 3*( | − 1) | |
| m |
| 21 | ||
log217=log21 | = 1−log213 =m ⇒ log213=1−m | |
| 3 |
| log2127 | 3log217 | 3(1−m) | ||||
log721= | = | = | ||||
| log217 | m | m |
| 3log213 | ||
log721=...... = | ||
| m |