Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna 1) log(3 x 4 ) + log( x + 2 ) = log(15 x + 2 ) + log x log 5
log(3 x 4 )( x + 2 ) = log 5
(15 x + 2)x5
cancelando los logaritmos tenemos.
(3x 4)(x + 2) = (15 x + 2)x5(3 x 4 )( x + 2 ) = (15 x + 2 )x
5[3 x 2 + 6 x 4 x 8] = 15 x 2 + 2 x
15 x 2 + 30 x 20 x 40 = 15 x 2 + 2 x10 x 2 x = 40 8 x = 40 x= 40 8 x=5
2) log 2 +
1 1 log( x 1) log 3 = log x log(2 x + 5) 2 21
2( x 1) 2 x log = log 1 3 (2 x + 5)2 2 x 1 x = / 3 2x + 5 3 2x + 5 6 x 1 2 x + 5 3x 2 x + 5 = 3 2x + 52
(x 1)(2 x + 5) = 3x /
2
4( x 1)(2 x + 5) = 9 x 2 4[2 x 2 + 5 x 2 x 5] = 9 x 2
8 x 2 + 20 x 8 x 20 = 9 x 2 8 x 2 9 x 2 + 12 x 20 = 0 x 2 + 12 x 20 = 0 / 1 x 2 12 x + 20 = 0 ( x 10 )( x 2 ) = 0y
x 10 = 0
x2=0
x1 = 10
x2 = 2
Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna 3)
1 1 =1 log x 3 log x + 2
Se tiene que multiplicar por
(log x 3)(log x + 2)
1 1 = 1 / (log x 3)(log x + 2 ) log x 3 log x + 2
(log x 3)(log x + 2) (log x 3)(log x + 2) = (log x 3)(log x + 2) (log x 3) (log x + 2)log x + 2 (log x 3) = (log x 3)(log x + 2 ) log x + 2 log x + 3 = (log x ) + 2 log x 3 log x 62
log x log x (log x ) 2 log x + 3 log x + 2 + 3 + 6 = 02
(log x ) + lox + 11 = 0 / 12
(log x )Sea y
2
log x 11 = 0
y = log x
y 2 y 11 = 0 hay que resolver la ecuacin cuadrtica en
y1, 2 = y1 =
1 1 4 1 11 1 1 + 44 1 45 = = 2 1 2 2y
y = log x
1 + 45 2
y2 =
1 45 2
Luego como
Tenemos que log x =
1 + 45 2
y
log x =
1 x 2
1 + 45 x1 = log 1 2 Encontrando que:
y
1 45 x2 = log 1 2
x1 = 7146,640995
y
x2 = 0,001399258
Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna
4) log( x + a ) + log( x b ) log x log( x a ) = 0
log( x + a )( x b ) = log x( x a )
(x + a )( x b ) = x( x a )x 2 bx + ax ab = x 2 ax bx + ax + ax = ab bx + 2ax = ab x(2a b ) = ab x=
ab 2a b
5) log x log( x a ) = log( x a ) log( x + a )
log
(x a ) x = log xa (x + a )2
x( x + a ) = ( x a )
x 2 + ax = x 2 2ax + a 2 ax + 2ax = a 2x(a + 2a ) = a2
a2 a x= = 3a 3
6) log(2 x + 1) + 2 log(3 x 4) 2 = 02
log(2 x + 1) (3 x 4 ) = 22 2
log(2 x + 1) (3 x 4 ) = log1002 2
[(2 x + 1)(3x 4)]
2
= 100
(2 x + 1)(3x 4) = 106 x 2 8 x + 3x 4 10 = 0 6 x 2 5 x 14 = 0
Ahora hay que resolver la ecuacin cuadrtica encontrada, entonces tenemos
x1, 2 =
5 25 4 6 14 5 25 + 336 5 361 5 19 = = = de donde 26 12 12 12
Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna
x1 =
5 + 19 24 = =2 12 12
y
x2 =
5 19 14 7 = = 12 12 6
7) log x 1 + log x + 4 log( x + 1) = 0
log x 1 x + 4 = log( x + 1)
( x 1)(x + 4) = x + 1x 2 + 4 x x 4 = x + 1/ 2Se eleva todo al cuadrado para eliminar la raz
x 2 + 3x 4 = x 2 + 2 x + 13x 2 x = 1 + 4 x=58) 2 log 7 x + 5 log(9 x 7 ) = 0
log 7 x + 5 = log(9 x 7 )2
(
)
log(7 x + 5) = log(9 x 7 ) 7x + 5 = 9x 7 7 x 9 x = 7 5 2 x = 12 / 1 x= 12 =6 2
9) log 3 x + 4 + log x + 5 = 1 + log1,2
log( 3 x + 4 x + 5 ) = log10 + log
12 10
log
(3x + 4)(x + 5) = log 10 1210
simplificando se obtiene
3 x 2 + 15 x + 4 x + 20 = 12 / 2
elevando a ambos lados al cuadrado queda
3x 2 + 19 x + 20 = 144 3x 2 + 19 x 124 = 0 19 19 2 4 3 124 19 361 + 1488 19 1849 x1, 2 = = = 23 6 6
Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna =
19 43 6
obteniendo para x
x1 =
24 =4 6
y
x2 =
62 31 = 6 6
log(5 + x 2 ) 10) =2 log(3 + x )
log(5 + x 2 ) = 2 log(3 + x ) log(5 + x 2 ) = log(3 + x )2
5 + x 2 = 9 + 6x + x 25 = 9 + 6x 5 9 = 6x 4 = 6x
4 2 x= = 6 3
log(x 3 56 ) 11) =3 log( x 2 ) log(x 3 56) = 3 log( x 2 ) log(x 3 56) = log( x 2 ) x 3 56 = ( x 2 )3 3
x 3 56 = ( x 2 )( x 2 )
2
x 3 56 = ( x 2 )(x 2 4 x + 4 )
x 3 56 = x 3 4 x 2 + 4 x 2 x 2 + 8 x 8 6 x 2 12 x 48 = 0 / : 6 x2 2x 8 = 0x1, 2 = 2 4 4 1 8 2 4 + 32 2 36 2 6 = = = 2 1 2 2 2donde
x1 =
2+6 8 = =4 2 25 x 2 + 3 x 92
y
x2 =
26 4 = = 2 2 2
12)
7
=1
Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna
7 5 x +3 x92 = 1 / log2
log 7 5 x +3 x 92 = log12
(7
5 x 2 + 3 x 92
)log 7 = 0
5 x 2 + 3x 92 = 0 3 32 4 5 92 3 9 + 1840 3 1849 3 43 x1, 2 = = = = 25 10 10 10Obtenindose:
x1 =
3 + 43 40 = =4 10 103 x +5
y
x2 =
3 43 46 = 4,6 = 10 10
13) (0,0625)
: (0,5)
x 7
=1
Aplicamos logaritmos a ambos lados del signo igual
(0,0625)
3 x +5
: (0,5)3 x +5
x 7
= 1 / logx 7
log (0,0625)
[
: (0,5)
] = log1
(0,0625) log (0,5)log(0,0625)
3 x +5
x 7
= log1 log(0,5)x 7
3 x +5
= log1
(3x + 5)log(0,0625) (x 7 )log(0,5) = log13 x log 0,625 + 5 log 0,0625 x log 0,5 + 7 log 0,5 = log1 3 x log 0,0625 x log 0,5 = log1 5 log 0,0625 7 log 0,5Factorizando en el lado izquierdo del signo igual, tenemos
x(3 log 0,0625 log 0,5) = log1 5 log 0,0625 7 log 0,5
x= x=
log1 5 log 0,0625 7 log 0,5 3 log 0,0625 log 0,5 (5 log 0,0625 + 7 log 0,5) = 2, 45 3 log 0,0625 log 0,5
Ecuaciones con logaritmos PROFESOR: Moiss Inostroza Cerna 14) 7 log x 2 log 35850 = 4 log
x 5
x7 x log = log 2 3585 5 x7 x = 35852 5 4
4
x 7 (3585) = x4 54 x3
2
(3585) =
2
(5 )
2 2
2 3585 x = x = 3 (143,4 ) = 3 20563,56 = 27,39677444 25 3
2