Question

Dados (log de 3 na base 24=x) e (log de 5 na base 24=y) , determine (log de 8 na base 45).
( a ) 1+x/ x-y  

( b ) 1-x/ 2x+ y  

( c ) 1+x/ 2x- 3y  

( d ) 1-x/ x-y  

( e ) 1+x/ y

Answer 1

X+y=log(24,5)+log (24,3)

X+y=log (24,15)

X+y=log (24,45/3)

X+y=log (24,45)-log (24,3)

X+y=log (24,45)-x

2x+y=log (24,45)

2x+y=1/log(45,24)

2x+y=1/log(45,3×8)

2x+y=1/(log(45,3)+log(45,8))

2x+y=1/((1/log(3,45))+log(45,8))

2x+y=1/((1/log(3,9×5)+log (45,8))

2x+y=1/((1/(log(3,3^2)+log (3,5))+log(45,8))

2x+y=1/((1/(2+y/x)+log (45,8))

log(45,8)+1/(2x+y)/x)=1/ (2x+y)

Log(45,8)+x/(2x+y)=1/(2x+y)

log (45,8)=1/(2x+y)-x/ (2x+y)

log (45,8)= (1-x)/(2x+y)

Obs:

Log (3,5)=y/x

y/x=log (24,5)/log (24,3)

Pela propriedade de mudança de base nos temos : y/x=log (5,3)

log (a,b)=1/log (b,a) (pela propriedade de mudança de base )

log (a,b)=log de b na base a

MUDANÇA DE BASE :log (a,b)=log (c,b)/log(c,a) , sendo c uma base qualquer de minha escolha

log(a,a)=1

Log (a,b^c)=clog(a,b)