| 3 | ||
Wykaż, że jeśli m=log169, to log128= | ||
| 2m+2 |
| 1 | ||
m = log16 9 = log24 92 = | *2 log2 3 = 0,5 log2 3 | |
| 4 |
| 1 | 1 | 1 | ||||
log12 8 = | = | = | = | |||
| log8 12 | log23 12 | 13( log2 3 + log2 4) |
| 3 | 3 | |||
= | = | |||
| log2 3 + 2 | 2 m + 2 |
| 2a | ||
A takie zadanie: Udowodnij, że jeżeli log312=a, to log212= | ||
| a−1 |
| log3 12 | 2 log3 12 | 2 a | ||||
log2 12 = | = | = | ||||
| log3 2 | 2 log3 2 | a − 1 |
