(i) Dado que log(4n)(40·rq(3)) = log(3n)45 encontrar n^3.

(ii) Sean a y b números positivos tales que

log(9)a = log(15)b = log(25)(a + 2b)

Encontrar el valor de b/a.

PD : log(x)y significa el logaritmo en base x de y.

Saludos.

(i) log(4n)(40·rq(3)) = log(3n)45 = x

(4n)^x = 40·rq(3)
(3n)^x = 45

Luego, (4/3)^x = 40·rq(3) / 45 = 8·rq(3) / 9 = (4/3)^(3/2) ==> x = 3/2

Entonces, (3n)^(3/2) = 45 ==> n^3 = 75

Saludos,

(i) log(4n)(40·rq(3)) = log(3n)45 = x

(4n)^x = 40·rq(3)
(3n)^x = 45

Luego, (4/3)^x = 40·rq(3) / 45 = 8·rq(3) / 9 = (4/3)^(3/2) ==> x = 3/2

Entonces, (3n)^(3/2) = 45 ==> n^3 = 75

Saludos,

(i) log(4n)(40·rq(3)) = log(3n)45 = x

(4n)^x = 40·rq(3)
(3n)^x = 45

Luego, (4/3)^x = 40·rq(3) / 45 = 8·rq(3) / 9 = (4/3)^(3/2) ==> x = 3/2

Entonces, (3n)^(3/2) = 45 ==> n^3 = 75

Saludos,

(ii) log(9)a = log(15)b = log(25)(a + 2b) = x

(9)^x = a
(15)^x = b ó
(25)^x = a+2b


(3^x)(3^x) = a
(3^x)(5^x) = b
(5^x)(5^x) = a+2b

Dividiendo la primera entre la segunda y la segunda entre la tercera :

(3^x) / (5^x) = a/b = b/(a+2b) ==> a^2 +2ab = b^2 ==>

==> 1 + 2(b/a) = (b/a)^2 ==> b/a = 1+sqrt(2)

(ii) log(9)a = log(15)b = log(25)(a + 2b) = x

(9)^x = a
(15)^x = b ó
(25)^x = a+2b


(3^x)(3^x) = a
(3^x)(5^x) = b
(5^x)(5^x) = a+2b

Dividiendo la primera entre la segunda y la segunda entre la tercera :

(3^x) / (5^x) = a/b = b/(a+2b) ==> a^2 +2ab = b^2 ==>

==> 1 + 2(b/a) = (b/a)^2 ==> b/a = 1+sqrt(2)

(ii) log(9)a = log(15)b = log(25)(a + 2b) = x

(9)^x = a
(15)^x = b ó
(25)^x = a+2b


(3^x)(3^x) = a
(3^x)(5^x) = b
(5^x)(5^x) = a+2b

Dividiendo la primera entre la segunda y la segunda entre la tercera :

(3^x) / (5^x) = a/b = b/(a+2b) ==> a^2 +2ab = b^2 ==>

==> 1 + 2(b/a) = (b/a)^2 ==> b/a = 1+sqrt(2)