Trzy liczby log 2 , log (2^{x} - 1) , log (2^{x} + 3) ustawione w tej kolejności tworzą ciąg arytmetyczny. Oblicz x.
( 2^x -1 musi być > 0
Ciąg jest arytmetyczny, zatem mamy
log (2^x - 1) - log 2 = log ( 2^x + 3) - log ( 2^x - 1)
log[(2^x - 1)/2 ] = log [ (2^x + 3)/(2^x - 1)]
zatem
(2^x - 1)/2 = (2^x + 3)/(2^x - 1)
( 2^x - 1)^2 = 2*(2^x + 3)
Podstawienie: y =2^x
Mamy
(y - 1)^2 = 2*(y + 3)
y^2 -2y + 1 = 2y + 6
y^2 - 4y - 5 = 0
delta = (-4)^2 - 4*1*(-5) = 16 + 20 = 36
p(delty) = 6
y1 = [ 4 - 6]/2 = -2/2 = - 1
y2 = [ 4 + 6 ]/2 = 10/2 = 5
2^x1 = - 1 - sprzeczność
2^x2 = 5
x2 = log 2 ( 5)
Odp. x = log 2 ( 5 )
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( 2^x -1 musi być > 0
Ciąg jest arytmetyczny, zatem mamy
log (2^x - 1) - log 2 = log ( 2^x + 3) - log ( 2^x - 1)
log[(2^x - 1)/2 ] = log [ (2^x + 3)/(2^x - 1)]
zatem
(2^x - 1)/2 = (2^x + 3)/(2^x - 1)
( 2^x - 1)^2 = 2*(2^x + 3)
Podstawienie: y =2^x
Mamy
(y - 1)^2 = 2*(y + 3)
y^2 -2y + 1 = 2y + 6
y^2 - 4y - 5 = 0
delta = (-4)^2 - 4*1*(-5) = 16 + 20 = 36
p(delty) = 6
y1 = [ 4 - 6]/2 = -2/2 = - 1
y2 = [ 4 + 6 ]/2 = 10/2 = 5
zatem
2^x1 = - 1 - sprzeczność
2^x2 = 5
x2 = log 2 ( 5)
Odp. x = log 2 ( 5 )
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