Ejercitación
Expresa cada logaritmo en forma exponencial.
I a. ln (x - 1) = 4 b. Iny=5
clog, 4 = 22
d.log() = -3
e. log, 1 = 0
f. log 0,1 = -1
por favor nesesito que me den respuesta a los que me den les daré corazón cinco estrellas y mis grasias y los seguiré ❤❤❤
Respuestas a la pregunta
En las preguntas donde escribi: OJO , son las respuestas dadas incorrectamente por el usuario anterior que respondio a la pregunta.
Propiedades a utilizar:
(1) Logb A + Logb B = Logb A.B
(2) Logb A - Logb B = Logb A/B
(3) Logb A = Logb B , si y solo si : A=B
(4) Logb A^n = n Logb A , y viseversa
"b" , es la base del logaritmo.
" ^ " , significa que el numero esta elevado a.
Ejm: 5^2 = 5² = 25
a) log x= log 5-log 3+log 11 log x = log5/3 + log 11 log x = log (5/3)(11) log x = log 55/3 x = 55/3a)logx=log5−log3+log11logx=log5/3+log11logx=log(5/3)(11)logx=log55/3x=55/3
B) log y=log 6+log 3-1/5 log 5 log y= log(6)(3) - log5^{1/5} log y = log (18) - log 5{1/5} log y = log (18/5^{1/5} ) y = 8/5^{1/5} y=8/ \sqrt[5]{5}B)logy=log6+log3−1/5log5logy=log(6)(3)−log5
1/5
logy=log(18)−log51/5logy=log(18/5
1/5
)y=8/5
1/5
y=8/
5
5
C) log z=3 log 2-1/2 log 4+log5
log z = log2³ - log4^1/2 + log5
(*) log 2³ = log 8 (*) 4^1/2 = 2
Entonces:
log z = log 8 - log 2 + log 5
log z = log 8/2 + log 5
log z = log 4 + log 5
log z = log 4.5
log z = log 20
z = 20
OJO:
[ Opcional ] : (a+b)^1/3 = Raiz cubica de (a+b)
(b+c)² = b² + 2bc + c²
OJO
E) log w=(1/3)(log a+1/4(log a+3 log c) log w = 1/3 (log a + 1/4 (log a + logc^3) log w = 1/3 (log a + 1/4 (log a.c^3) log w = 1/3 ( log a + log (a.c^3)^{1/4} log w = 1/3 (log a.a^{1/4} .c^{3/4} ) log w = log [(a^{5/4})(c^{3/4})]^{1/3} w = a^{5/12} . c^{(3/4)(1/3)} w = a^{5/4} . c^{1/4} w= \sqrt[4]{a^5 . c}E)logw=(1/3)(loga+1/4(loga+3logc)logw=1/3(loga+1/4(loga+logc
3
)logw=1/3(loga+1/4(loga.c
3
)logw=1/3(loga+log(a.c
3
)
1/4
logw=1/3(loga.a
1/4
.c
3/4
)logw=log[(a
5/4
)(c
3/4
)]
1/3
w=a
5/12
.c
(3/4)(1/3)
w=a
5/4
.c
1/4
w=
4
a
5
.c
OJO:
F) log x =- (log a . log b - log a . log b . logc] log(x) = - ( loga^{logb} - loga^{logb.logc} log(x) = - (log a^{logb} / a^{log b.log c} ) log (x) = - log a^{logb - logb.logc log(x) = -1 log a^{(logb)(1-logc) log(x) = log [a^{(logb)(1-logc)]^{-1 log (x) = log a^{(logb)(logc-1) x=a^{logb(logc-1)