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

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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) About Me  (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
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+.HENCE
x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+
x+=+%282+%2B-+sqrt%28+4%2B4+%29%29%2F2+
x+=+%282+%2B-+sqrt%28+8+%29%29%2F2+
x+=+%282+%2B-2%2Asqrt%282%29%29%2F2+
x+=+%281+%2Bsqrt%282%29%29+AS X CANNOT BE NEGATIVE FOR LN X TO BE REAL