Question

1- a) resolva a equação 10^1-x=0,0001b) log (2x-3)+log (x+2)=2. log x 2) sabendo que log 5479=3,45347 e que log 1235= 1,5678, calcule o valor de: a) log 547,9 

1- a) resolva a equação 10^1-x=0,0001 
b) log (2x-3)+log (x+2)=2. log x 

2) sabendo que log 5479=3,45347 e que log 1235= 1,5678, calcule o valor de: 

a) log 547,9 
b)log 1235,5

Answer 1

1.a)
10¹-x=0,0001 
10¹-x= 10^-4 --> x=10¹ +10^-4= 10,0001

1.b)
log(2x-3)+ log(x+2) = 2log(x)
log((2x-3).(x+2)) = log(x²)
log(2x²+4x-3x-6) = log(x²)
log(2x² +x-6) = log(x²)

2x² +x-6=x² -->x² +x-6=0
DElta--> 1-4.1.-6= 25
-1+/- 5/2
X1=2 e X2=-3

log((2x-3).(x+2)) = log(x²)
log((2.2-3).(2+2)) = log(2²)
log((1).(4)) = log(4)
log(4)=log(4) --> certo

log((2.-3-3).(-3+2)) = log(-3²)
log((-9).(-1)) = log(9) --> errado

2.a)
log 547,9 
 547,9=  5479/10, logo:
log 5479/10
log 5479/10 =Log5479- log10 = 3,45347- 1= 2,45347

2.b)
log 1235,5
1235,5= 12355/10, logo:
log 12355/10
log 12355/10= Log12355 - Log 10= 1,5678 - 1= 0,5678