• Matéria: Matemática
  • Autor: Tarcisiodt
  • Perguntado 8 anos atrás

Sabendo que log 2=0,301 e log3=0,477 cálcule as questões A= log 7,2 B=0,72 e C=0,072

Respostas

respondido por: baebergamota
0
7,2=72/10
0,72=72/100
0,072=72/1000
72=2.2.2.3.3
72:2^3.3^2

log7,2=
log72/10=
log2^3.3^2-log10=
log2^3+log3^2-log10=
3log2+2log3-log10=
3.0,301+2.0,477-1=
0,903+0,954-1=
1,857-1=
0,857

log0,72=
log72/100=
log2^3.3^2/10^2
log2^3+log3^2-log10^2=
3log2+2log3-2log10=
3.0,301+2.0,477-2.1=
0,903+0,954-2=
1,857-2=
-0,143

log0,072=
log72/1000=
log2^3.3^2/10^3=
log2^3+log3^2-log10^3=
3log2+2log3-3log10=
3.0,301+2.0,477-3.1=
0,903+0,954-3=
1,857-3=
-1,143



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