| 4a−2 | ||
Wykaż, że jeżeli log1612=a, to log243= | ||
| 4a+1 |
| 4a − 2 | 4log16 12 − 2 | |||
P = | = | = | ||
| 4a + 1 | 4log16 12 + 1 |
| log2 12 − 2 | log2 3 | |||
= | = | = log24 3 = L | ||
| log2 12 + 1 | log2 24 |
| log212−log24 | 4a−2 | |||
log243=U{log23}{log224= | = | |||
| log212+log22 | 4a+1 |
| 4log1612−2 | log212−2 | ||
=> | |||
| 4log1612+1 | log212+1 |
| 1 | ||
logac b = | loga b | |
| c |
| |||||||||||
P= | = | ||||||||||
|
| log2(12:4) | |||||||||||
= | = | |||||||||||
| log2(12*2) |