se log 2= 0,3 , log 3= 0,5
log 5= 0,7 e log 7= 0,85 , responda:
(exercicios na foto)
Soluções para a tarefa
Resposta:
Vide abaixo
Explicação passo-a-passo:
a)
log14 49 =
log14 7^2 =
2. log14 7 =
2. (log 7 / log 14) =
2. (log 7 / log 7. 2) =
2. [ log 7 / (log 7 + log 2) ] =
2. [0,85 / (0,85 + 0,3)] =
2. 0,85/ 1,15 =
~ 1,478
b)
log12 64 =
log12 8^2 =
2. log12 8 =
2. log12 2^3 =
2.3. log12 2 =
6. log12 2 =
6. (log 2 / log 12) =
6. (log 2 / log 4.3) =
6. [ log 2 / (log 4 + log 3) ] =
6. [ log 2 / (log 2^2 + log 3) ] =
6. [ log 2 / (2.log 2 + log 3) ] =
6. [ 0,3 / (2. 0,3 + 0,5) ] =
6. 0,3 / 1,1 =
~ 1,636
c)
log21 40 =
log21 8.5 =
log21 8 + log21 5 =
log21 2^3 + log21 5 =
3. log21 2 + log21 5 =
3. (log 2 / log 21) + (log 5 / log 21) =
3. (log 2 / log 7. 3) + (log 5 / log 7. 3) =
3. (log 2 / (log 7 + log 3)) + (log 5 / (log 7 + log 3)) =
3. (0,3 / (0,85 + 0,5)) + (0,7 / (0,85 + 0,5)) =
0,9/ 1,35 + 0,7/ 1,35 =
1,6/ 1,35 =
~ 1,185
d)
log0,05 0,03 =
log0,05 3/100 =
log0,05 3 - log 0,05 100 =
log 3 / log 0,05 - log 100 / log 0,05 =
log 3 / log 5/100 - 2 / log 5/100 =
log 3 / (log 5 - log 100) - 2 / (log 5 - log 100) =
0,5 / (0,7 - 2) - 2 / (0,7 - 2) =
(0,5 - 2) / (0,7 - 2) =
-1,5 / -1,3 =
~ 1,154
Blz?
Abs :)