| 1 | 25√64 | |||
log | ||||
| 4 | √8 |
| 1 | ||
rozkładałem to na ( | )x = 2 do pot. 11/5 / 23 | |
| 4 |
| 1 | ||
Odpowiedź jest −7/20 a mi wychodzi | do potegi x = 2 do pote. −4/5 | |
| 4 |
| 4 | ||
czyli (2−2)x=2− | ||
| 5 |
| 1 | |
= 2−2 | |
| 4 |
| 25√64 | 21+6/5 | ||
= | = 211/5 − 3/2 = 27/10 | ||
| √8 | 23/2 |
| 1 | 7 | 7 | ||||
.... = − | * | log22 = − | ||||
| 2 | 10 | 20 |
| 11 | 3 | 3 | ||||
2 do potęgi | − | , bo 1 nad √8 to 2 do − | ||||
| 5 | 2 | 2 |