Question

1) Dados log2= 0,301, log3=0,477, log5= 0,699, calcule:


a) log 7√2,5:

b)log 0,001:


2)Resolva as seguintes equações:


a)log x-3(2x-6)=2

b)log(log2(log3 3x))=0

c)log27[2+log5(x²-4)]=1/3

Answer 1

1

A)

Log7√25/10

Log7√5/2

Log[7.(5/2)^1/2]

Log7+(1/2.[Log5-Log2])

0,84+(0,699-0,301)/2

084+0,199 = 1,039

B) log0,001

Log(1/1000)

Log1-Log1000

-log10³

-3log10

-3