| 1 | ||
mamy wzór logar (b) = | loga (b), zastosujmy go: | |
| r |
| 1 | 1 | |||
= − | log2 (27) + | log2 (27), teraz skorzystajmy ze wzoru: loga (br) = rloga (b) | ||
| 3 | 2 |
| 3 | 1 | 5 | ||||
w = −log2(3) + | log2(3) = | log2(3) ≠ |  | |||
| 2 | 2 | log3(2) |
| 1 | ||
w = − | log2(27) + 2log2(27) = −log2(3) + 6log2(3) = 5log2(3) | |
| 3 |
| 1 | ||
stosujemy wzór: loga(b) = | i mamy ostatecznie: | |
| logb(a) |
| 5 | ||
w = | | |
| log3(2) |
| 1 | 2 | 1 | ||||
log1/827= | i log√227= | i log272= | log32 | |||
| −3log272 | log272 | 3 |
| −1 | 6 | |||
W= | + | |||
| 3log272 | 3log272 |
| 5 | ||
W= | ||
| log32 |