Question

se log 2= 0,3 , log 3= 0,5
log 5= 0,7 e log 7= 0,85 , responda:

(exercicios na foto)​

Answer 1

Resposta:

e) 1,5

f) ~ 0,933

g) ~ 0,659

h) 6

i) 0,6

Explicação passo-a-passo:

e)

log20 84=

log20 (2^2).3.7=

2.log20 2 + log20 3 + log20 7=

2.(log2 / log 20) + (log 3 / log 20) + (log 7 / log 20) =

2.(log2 / log 5. 2^2) + (log 3 / log 5. 2^2) + (log 7 / log 5. 2^2) =

2.(log2 / (log 5 + 2.log 2)) + (log 3 / (log 5 + 2.log 2)) + (log 7 / (log 5 + 2.log 2)) =

2.(0,3 / (0,7 + 2. 0,3)) + (0,5 / (0,7 + 2. 0,3)) + (0,85 / (0,7 + 2. 0,3)) =

0,6/1,3 + 0,5/1,3 + 0,85/1,3 =

1,95/1,3 = 1,5

f)

log32 25=

log32 5^2

2.log32 5=

2.(log 5 / log 32)=

2.(log 5 / log 2^5)=

2.(log 5 / 5.log 2)=

2.(0,7 / 5. 0,3)=

1,4/ 1,5=

~ 0,933

g)

log160 28=

log160 2^2. 7=

log160 2^2 + log160 7=

2.log160 2 + log160 7=

2.(log 2 / log 160) + (log 7 / log 160)=

(2.log 2 + log 7) / log 160=

(2.log 2 + log 7) / log 2^5. 5=

(2.log 2 + log 7) / (5.log 2 + log 5)=

(2. 0,3 + 0,85) / (5. 0,3 + 0,7)=

1,45/2,2=

~ 0,659

h)

lograiz(3) 32=

lograiz(3) 2^5=

5.lograiz(3) 2=

5.(log 2 / log raiz(3))=

5.(log 2 / log 3^(1/2))=

5.(log 2 / (1/2).log 3)=

5.(0,3 / 0,5.0,5)=

1,5/0,25= 6

i)

log3raiz(3) 2.raiz(2)=

log 2.raiz(2) / log 3.raiz(3)=

(log 2 + log raiz(2)) / (log 3 + log raiz(3))=

(log 2 + log 2^(1/2)) / (log 3 + log 3^(1/2))=

(log 2 + 0,5. log 2) / (log 3 + 0,5. log 3)=

(0,3 + 0,5. 0,3) / (0,5 + 0,5. 0,5)=

0,45/0,75= 0,6

Blz?

Abs :)