funkcja wykładnicza i logarytmiczna
xyz: a) 25log51/2+1/2*8log23√3−1/3
b) 8√(3√3−1)√3+1*3√9
6 sty 15:53
Godzio:
| | 1 | | 5 | |
25log5(1/2) + 1/2 = 52log5(1/2) + 1 = 5{log5(1/4) * 5 = |
| * 5 = |
| |
| | 4 | | 4 | |
| | 3 | |
8log23√3 − 1/3 = 23log23√3 − 1 = U{2log23{2} = |
| |
| | 2 | |
(3
√3 − 1)
√3 + 1 *
3√9 = 3
3 − 1 *
3√9 = 9 * 9
1/3 = 9
4/3
8√94/3 = 9
4/24 = 3
8/24 = 3
1/3 =
3√3
6 sty 16:04
Michal: 25log3 5−2
15 cze 09:23