Quais as relações entre a, b, c ∈ R*₊ , com a , b , c ≠ 1, a ≠ c e b ≠ 1/c para que a equação abaixo seja válida
loga(bc)β.logb(a/c)α.logcb = αβ[A - B + C - D],
em que
- A=logc(a) . loga(b) . logb(a)
- B=loga(b) . logb(c) . logc(a)
- C=loga(c) . logc(a) . logb(a)
- D=loga(c) . logb(c) . logc(a), tal que α,β ∈ R*.
- a + b = c
- a - b = 0
- a + b + c = 1
- c - 2b = 0
- 2a + 3c = 1
Resolução
Sabemos que uma das propriedades dos logaritmos é:
logba × logab = 1
Então:
A = logca . logab . logba
A = logca
B = logab . logbc . logca
B = (log b / log a) . (log c / log b) . (log a / log c) (mudança de
base)
B = 1
C = logac . logca . logba
C = logba
D = logac . logbc . logca
D = logbc
Voltando pra equação pedida:
loga(bc)β.logb(a/c)α.logcb
= αβ[A - B + C - D]
β.loga(bc).α.logb(a/c).logcb = αβ[A - B
+ C - D]
loga(bc).logb(a/c).logcb = A - B + C -
D
loga(bc).logb(a/c).logcb =
logca - 1 + logba - logbc
loga(bc).logb(a/c).logcb =
logca/c + logba/c
loga(bc).logb(a/c).logcb =
(logba/c / logbc) + logba/c
loga(bc).logb(a/c).logcb =
logba/c (1/logbc + 1)
loga(bc).logcb = (1/logbc + 1)
loga(bc).logcb = logcb + 1
loga(bc) = 1 + 1/logcb
loga(bc) = 1 + logbc
loga(bc) = logbbc
loga(bc) = logabc / logab
1 = 1 / logab
logab = 1
a = b
a - b = 0
Gabarito: (B)