SOLUTION: How would i solve this natural log in this equation: ln(x+1)=ln(3x+1)-ln(x)...thanks

Question 24553: How would i solve this natural log in this equation: ln(x+1)=ln(3x+1)-ln(x)...thanks
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
ln(x+1)=ln(3x+1)-ln(x)..
USING LN (A*B)=LN A + LN B ....AND
LN (A/B)=LN A - LN B......WE GET
LN (X+1)=LN {(3X+1)/X}
TAKING ANTI LOGS...
X+1=(3X+1)/X
X(X+1)=3X+1
X^2+X=3X+1
X^2+X-3X-1=0
X^2-2X-1=0
THE SOLUTION OF AX^2+BX+C=0 IS GIVEN BY
.HENCE

AS X CANNOT BE NEGATIVE FOR LN X TO BE REAL
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