Question

sendo log 2 = x e log 3 = y , calcule :
a. log2 432
b. log4 20
c. log2 540
d. log3 540

Answer 1

Fatorando os números:
432 = 2.2.2.2.3.3.3 = (2^4).(3^3)
540 = 2.2.3.3.3.5 = (2^2).(3^3).5

a) log2 432 = log2 (2^4).(3^3) = log2 (2^4) + log2 (3^3) = 4.(log2 2) + 3.(log2 3) =
4 + 3(log 3/log 2) = 4 + 3.y/x = (4x + 3y)/x

b) log4 20 = log 20/log 4 = log (2.10)/log (2^2) = (log 2 + log 10)/(2.log 2) =
(x + 1)/(2.x)

c) log2 540 = log2 [(2^2).(3^3).5] = log2 [2.(3^3).10] = log2 2 + log2 (3^3) +
 log2 10
= 1 + 3.(log2 3) + log2 10 = 1 + 3.(log 3/log 2) + (log 10/log 2) =
1 + 3.y/x + 1/x = (x + 3y + 1)/x

d) log3 540 = log3 [(2^2).(3^3).5] = log3 [2.(3^3).10] = log3 2 + log3 (3^3) +
 log3 10
= (log 2/log 3) + 3.(log3 3) + (log 10/log 3) = 1 + 3.(log 2/log 3)
+ (log 10/log 2)
= x/y + 3 + 1/y = (x + 3y +1)y