| 3(1−m) | ||
469. Niech m=log217. Wykaż, że log727= | . | |
| m |
| 3log3 | log21 | log3 + log7 | ||||
m = | oraz log727 = | = | ||||
| log7 | log7 | log7 |
| log3 + log7 | 3log3 | |||
m = | oraz log727 = | |||
| log7 | log7 |
| log7 | log7 | |||
m=log217= | = | ? | ||
| log21 | log3+log7 |
| log77 | 1 | 1 | ||||
log217= | = | =m ⇒1+log73= | ⇒ | |||
| log721 | 1+log73 | m |
| 1 | 1−m | |||
log73= | −1= | |||
| m | m |
| 3*(1−m) | ||
log727=3log73= | ||
| m |