TEKST ZADATKA
Dokazati da, ako je c 2 − b 2 = a 2 , c^2 - b^2 = a^2 , c 2 − b 2 = a 2 , c − b ≠ 1 , c - b \neq 1 , c − b = 1 , c + b ≠ 1 , c + b \neq 1 , c + b = 1 , a > 0 , a > 0 , a > 0 , b > 0 , b > 0 , b > 0 , c > 0 , c > 0 , c > 0 , važi jednakost
log c + b a + log c − b a = 2 log c + b a ⋅ log c − b a . \log_{c+b} a + \log_{c-b} a = 2 \log_{c+b} a \cdot \log_{c-b} a . log c + b a + log c − b a = 2 log c + b a ⋅ log c − b a .
REŠENJE ZADATKA
Razmotrimo prvo slučaj kada je a = 1. a = 1 . a = 1. Tada su svi logaritmi oblika log x 1 , \log_x 1 , log x 1 , što je jednako nuli, pa za levu stranu jednakosti važi:
log c + b 1 + log c − b 1 = 0 + 0 = 0 \log_{c+b} 1 + \log_{c-b} 1 = 0 + 0 = 0 log c + b 1 + log c − b 1 = 0 + 0 = 0 Prikaži korak 1 Desna strana za a = 1 a = 1 a = 1 je takođe jednaka nuli, pa jednakost trivijalno važi.
2 log c + b 1 ⋅ log c − b 1 = 2 ⋅ 0 ⋅ 0 = 0 2 \log_{c+b} 1 \cdot \log_{c-b} 1 = 2 \cdot 0 \cdot 0 = 0 2 log c + b 1 ⋅ log c − b 1 = 2 ⋅ 0 ⋅ 0 = 0 Prikaži korak 2 Pretpostavimo sada da je a ≠ 1. a \neq 1 . a = 1. Primenjujemo pravilo za promenu osnove logaritma log y x = 1 log x y \log_y x = \frac{1}{\log_x y} log y x = l o g x y 1 na levu stranu jednakosti kako bismo prešli na osnovu a . a . a .
log c + b a + log c − b a = 1 log a ( c + b ) + 1 log a ( c − b ) \log_{c+b} a + \log_{c-b} a = \frac{1}{\log_a (c+b)} + \frac{1}{\log_a (c-b)} log c + b a + log c − b a = log a ( c + b ) 1 + log a ( c − b ) 1 Prikaži korak 3 Svodićemo dobijeni izraz na zajednički imenilac.
1 log a ( c + b ) + 1 log a ( c − b ) = log a ( c − b ) + log a ( c + b ) log a ( c + b ) ⋅ log a ( c − b ) \frac{1}{\log_a (c+b)} + \frac{1}{\log_a (c-b)} = \frac{\log_a (c-b) + \log_a (c+b)}{\log_a (c+b) \cdot \log_a (c-b)} log a ( c + b ) 1 + log a ( c − b ) 1 = log a ( c + b ) ⋅ log a ( c − b ) log a ( c − b ) + log a ( c + b ) Prikaži korak 4 Primenjujemo pravilo za zbir logaritama sa istom osnovom: log a x + log a y = log a ( x y ) \log_a x + \log_a y = \log_a (xy) log a x + log a y = log a ( x y ) na brojilac.
log a ( c − b ) + log a ( c + b ) = log a ( ( c − b ) ( c + b ) ) = log a ( c 2 − b 2 ) \log_a (c-b) + \log_a (c+b) = \log_a ((c-b)(c+b)) = \log_a (c^2 - b^2) log a ( c − b ) + log a ( c + b ) = log a (( c − b ) ( c + b )) = log a ( c 2 − b 2 ) Prikaži korak 5 Iz uslova zadatka imamo da je c 2 − b 2 = a 2 . c^2 - b^2 = a^2 . c 2 − b 2 = a 2 . Zamenjujemo ovo u dobijeni izraz za brojilac.
log a ( c 2 − b 2 ) = log a ( a 2 ) \log_a (c^2 - b^2) = \log_a (a^2) log a ( c 2 − b 2 ) = log a ( a 2 ) Prikaži korak 6 Koristeći osobinu logaritma log a x s = s log a x \log_a x^s = s \log_a x log a x s = s log a x i log a a = 1 , \log_a a = 1 , log a a = 1 , računamo vrednost brojioca.
log a ( a 2 ) = 2 log a a = 2 ⋅ 1 = 2 \log_a (a^2) = 2 \log_a a = 2 \cdot 1 = 2 log a ( a 2 ) = 2 log a a = 2 ⋅ 1 = 2 Prikaži korak 7 Vraćamo dobijenu vrednost brojioca u razlomak.
2 log a ( c + b ) ⋅ log a ( c − b ) \frac{2}{\log_a (c+b) \cdot \log_a (c-b)} log a ( c + b ) ⋅ log a ( c − b ) 2 Prikaži korak 8 Ovaj izraz možemo zapisati kao proizvod, a zatim ponovo primeniti pravilo za promenu osnove logaritma kako bismo se vratili na početne osnove.
2 ⋅ 1 log a ( c + b ) ⋅ 1 log a ( c − b ) = 2 log c + b a ⋅ log c − b a 2 \cdot \frac{1}{\log_a (c+b)} \cdot \frac{1}{\log_a (c-b)} = 2 \log_{c+b} a \cdot \log_{c-b} a 2 ⋅ log a ( c + b ) 1 ⋅ log a ( c − b ) 1 = 2 log c + b a ⋅ log c − b a Prikaži korak 9 Dobili smo desnu stranu početne jednakosti, čime je dokaz završen.
log c + b a + log c − b a = 2 log c + b a ⋅ log c − b a \log_{c+b} a + \log_{c-b} a = 2 \log_{c+b} a \cdot \log_{c-b} a log c + b a + log c − b a = 2 log c + b a ⋅ log c − b a Prikaži korak 10