(log52 2 − 1) * log2 5 | 2 | ||
= log | |||
log2 5 + log5 2 + 2 | 5 |
a * (a2 − 1) * 1/a | a2 − 1 | a − 1 | ||||
L = | = | = | = | |||
a * (1/a + a + 2) | (a+1)2 | a+ 1 |
log5 2 − 1 |
| 2 | |||||||||||||
= | = | = log | = P | ||||||||||||
log5 2 + 1 | log5 10 | 5 |
log25 | 1 | ||
= | |||
log25+log52+2 | log522+2log52+1 |
log522−1 | log52−1 | 2 | 2 | ||||
= | =log10 | =log | |||||
log522+2log52+1 | log52+1 | 5 | 5 |
1 | ||
Wówczas log25 = | ||
x |
|
| ||||||||||||||||||
lewa strona = | = | = | |||||||||||||||||
|
|
x2−1 | (x−1)(x+1) | x−1 | log52−1 | |||||
= | = | = | = | = | ||||
1+x2+2x | (x+1)2 | x+1 | log52+1 |
log52−log55 |
| 2 | |||||||||||||
= | = | = log | = P | ||||||||||||
log52+log55 | log510 | 5 |
| 2 | ||||||||||||
Ostatnie przejście, z | do log | ||||||||||||
log510 | 5 |
logac | ||
wynika z zastosowania wzoru na zamianę podstawy logarytmu | = logbc dla a=5, | |
logab |
2 | ||
b=10, c= | ||
5 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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