Question

(log_3X)² - 5log_9(X)+1=0
Bases diferentes: 3 e 9

Answer 1

Resposta:

Se for assim   (log₃ x)² - 5log₉ x + 1 = 0

(log₃ x)² - 5log₉ x + 1 = 0

(log x/log 3)² - 5log x/log 9 + 1 = 0

(log x/log 3)² - 5log x/log 3² + 1 = 0

(log x/log 3)² - 5log x/2log 3 + 1 = 0

(log x/log 3)² - (5/2)log x/log 3 + 1 = 0

Fazendo y =log x/log 3 = log₃  x

y²-(5/2)* y+1=0

y=1/2 = log₃  x  ==>x=3^(1/2) =√3

y=2= log₃  x  ==>x=3²=9