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Cómo resuelvo estos logaritmos?
3^2x-1=64 (tres elevado a la dos equis menos uno es igual a 64)
5^3x=25^x-3 (cinco elevado a la tres equis es igual a 25 elevado a la equis menos 3)
log3y=2
log64x=1/3--(un tercio)
log12 8=n
Muchas gracias a quienes me ayuden!!!!!!!
si me pudieran explicar paso por paso por que nada mas no entiendo
2 Answers
- 1 decade agoFavorite Answer
3^(2x - 1) = 64 aplicás log decimal en ambos miembros
log 3^(2x - 1)= log 64
(2x - 1) log 3 = log 64
2x - 1 = log 64/ log 3
2x - 1 = 1,806 / 0,477
2x = 3,786 +1
x = 4,786 / 2
x = 2,393 si hacés la verificación te va a dar 64,02 por los decimales que descartamos
.............................................
5^(3x) = 25^(x-3)
aplicamos log base 5
log 5^(3x) = log 25^(x-3)
5 5
3x log 5 = (x-3) log 25
5 5
3x * 1 = (x-3) * 2 aplico prop. distributiva
3x = 2x - 6
3x- 2x = -6
x = -6
................................................
log 3y = 2 significa que
10 ^2 = 3y
100/3 = y
...........................................
log64x=1/3
10 ^(1/3) = 64 x
2,154 / 64 = x
0,033 = x
.............................................
log12 8=n
2,107 =n
Suerte
- 1 decade ago
(2x-1)ln3 = 3^(2x-1) = 64
(2x-1)ln3=64
2x=64/ln3 - 1
x= 32/ln3 -1/2
5^3x=25^(x-3)=5^(2*(x-3))
3x(ln5)=(2x-6)ln5
3x=2x-6
x=-6
log3y=2
3y=e^2
y=(e^2)/3
log64x=1/3
64x=e^(1/3)
x=(e^(1/3))/64
log12 8=n en esta que quieres saber
