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# Clase 123 log2(x – 3) + log2 x = 5 log6(x2 – 4) - log6 2(x + 2) = 2

## Presentación del tema: "Clase 123 log2(x – 3) + log2 x = 5 log6(x2 – 4) - log6 2(x + 2) = 2"— Transcripción de la presentación:

Clase 123 log2(x – 3) + log2 x = 5 log6(x2 – 4) - log6 2(x + 2) = 2
Ejercicios sobre ecuaciones logarítmicas log2(x – 3) + log2 x = 5 log6(x2 – 4) - log6 2(x + 2) = 2

logab = x ssi ax = b (a > 0, a  1, b > 0)
Definición de logaritmo logab = x ssi ax = b (a > 0, a  1, b > 0) loga(b·c) = logab + logac (c>0) loga = logab – logac (c 0) b c Propiedades logabx= x logab logab logcb logca = log b = logab ax 1 x

Ejercicio Para qué valores de x se satisfacen las siguientes ecuaciones: a) log3(x2–6x+1)–log3(x–3)= –1 b) log( x + 3) = 2 log 2x – 3 c) log2  x log2 x = 1

a) log3(x2–6x+1)–log3(x–3)= –1
=3 –1 x2–6x+1 x–3 = x2–6x+1 x–3 1 3 3x2–18x+3 = x–3 3x2–19x+6 = 0

3x2–19x+6 = 0 (3x – 1)(x – 6) = 0 3x – 1 = 0 ó x – 6 = 0 x1= 1 3
N.S. indefine los logaritmos MI: log3(x2–6x+1)–log3(x–3) = log3(36–36+1)–log3(6–3) = log31 –log3 3 MD: – 1 = 0 – 1 = – 1 Comp: – 1 = – 1

b) log( x + 3) = 2 log 2x – 3 log( x + 3) = log( 2x – 3 )2
 x + 3 = 2x – 3 2 2  x = 2x – 6 x = 4x2 – 24x + 36 4x2 – 25x + 36 = 0 (4x – 9)(x – 4) = 0 9 4 x1= ó x2 = 4

Comprobación para x = 4 M.I: log( x + 3) = log( 4 + 3) = log(2 + 3) = log 5 M.D: 2 log 2x – 3 = 2 log 2·4 – 3 = 2 log 8 – 3 = 2 log 5 = log( 5 )2 = log 5 Comparación: log 5 = log 5

c) log2  x + 3 + log2 x = 1 1 3 0 – 4 log2 [( x + 3 ) (x)] = 1 1 1 4
– 4 log2 [( x + 3 ) (x)] = 1 1 1 4 4 1 4 ( x + 3 ) (x) = 2 4 1 (x – 1)(x + 2)2 ( x + 3 )2 (x)2 = 4 x2(x+3) = 4 x3+3x2 – 4 = 0 = 0 x1= 1 ó x2 = – 2 N.S.

Para el estudio individual
Resuelve: a) log x + log (x+3) + 4 = 5 b) 2 log x = log 2 + log (x + 4) log  x+5 + x = 2 7 c) d) log2(3x+1)+log2(3x–2) = log 4 log 2 Resp: a) b) 4 c) d) 1

– 4 1 1 4 4 1 4 4 (x – 1)(x + 2)2

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