Find the value of x which satisfies the relation
Log10 3+log10 (4x+1)=log10 (x+1)+1

• log10 3+log10 (4x+1)=log10 (x+1)+1
  Log10 3+log10 (4x+1)=log10 (x+1)+log10 10
  Log10 (3(4x+1))=log10 (10(x+1))
  =3(4x+1)=10(x+1)=12x+3
  =10x+10
  =2x=7=x=7/2
• 12 years agoHelpfull: Yes(35) No(2)
• 10 is base==>>3(4x+1)=10(x+1)==>x=7/2
• 12 years agoHelpfull: Yes(6) No(0)
• [log10a + log10b = log10 ab]
  and also 1 can be written as log10 10,
  so question can be rewritten as: log10 (3(4x+1))=log10(10(x+1)),
  this can be solved to get 12x+3=10x+10,
  that gives x=7/2=3.5
• 12 years agoHelpfull: Yes(4) No(1)
• we know that loge e=1, so log10 10 =1
  then it becomes Log10 3+log10 (4x+1)=log10 (x+1)+log10 10
  log10 3 - log10 10=log10 (x+1)-log10 (4x+1)
  we know loga-logb =log(a/b)
  then log(3/10) 10= log(x+1/4x+1)10
  when bases are equal we can equate its powers
  then 3/10=x+1/4x+1
  12x+3=10x+10
  2x=7
  x=7/3
• 12 years agoHelpfull: Yes(1) No(7)