| 1 | ||
Skorzystajmy ze wzoru: logar (b) = | loga (b), a≠1 ∧ a>0 ∧ b>0 | |
| r |
| 1 | 1 | r | |||||||||
= | = | , zatem mamy: | |||||||||
| logar (b) |
| loga (b) |
| 1 | 1 | 1 | 1 | |||||
L = | + | + | + | |||||
| loga (b) | loga2 (b) | loga3 (b) | loga10 (b) |
| 1+2+3+...+10 | 55 | |||
L = | = | = P | ||
| loga (b) | loga (b) |
| 1 | 1 | 1 | |||
+ | + ... + | ||||
| logab | loga2b | loga10b |
| 55 | ||
= | ||
| logab |
(2)
|DB|=13, |AB|=a, |CD|=b, |DE|=2r, L=48=a+b+2c
Okrąg wpisany w trapez, zatem mamy: a+b=2c ⇒ L=4c=48 ⇒ c=12 ⇒ a+b=24
W ΔDEB:
| a+b | 24 | 5 | ||||
|EB|= | = | =12, zauważmy, że ΔDEB jest trójkątem (5,12,13) ⇒ 2r=5 ⇒ r= | ||||
| 2 | 2 | 2 |
| 12 | 78 | ||||||||
= 2R ⇒ R= | |||||||||
| 5 |