SOLUTION: Hi,I hope someone can answer my log question: log(x+2)-logx= log (x+5) Solve for x. Thanks alot!

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Question 4711: Hi,I hope someone can answer my log question:
log(x+2)-logx= log (x+5) Solve for x.
Thanks alot!

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
log(x+2)-logx=(x+5)

Use the second law of logarithms log+M+-+log+N+=+log+%28M+%2F+N%29+to express the left side as a quotient:
log+%28%28x%2B2%29%29+-+log+x+=+log+%28%28x%2B5%29%29+
log+%28%28x%2B2%29%2F+%28x%29%29+=+log+%28%28x%2B+5%29%29+

log M = log N means that M=N.
Therefore +%28x%2B2%29+%2F+x+=+%28x%2B5%29+

Multiply both sides of the equation by x:
x%2B+2+=+x%28x%2B+5%29+
x%2B2+=+x%5E2+%2B+5x+

Set equal to zero:
+0+=+x%5E2+%2B+5x+-+x+-+2
+0+=+x%5E2++%2B+4x+-+2

Are you sure you copied this right? They "usually" factor at this point. Nevertheless,
Solve by quadratic formula or completing the square. Completing the square is actually easier.
x%5E2++%2B+4x+-+2+=+0 Add +2 to each side and place a blank space on each side of the equation:
+x%5E2+%2B+4x+%2B+_____+=+2+%2B+_____

Take half of the x coefficient, and square it. Half of 4 is 2, and 2^2 = 4. Add +4 to each side of the equation on the blank lines above:
+x%5E2+%2B+4x+%2B+4+=+2+%2B+4
+%28x%2B+2%29+%5E2+=+6

Take square root of each side of the equation:
x%2B2+=+0+%2B-+sqrt+%286%29+

Subract 2 from each side:
x=+-2+%2B-+sqrt+%286%29

Of course, you must reject the value -2+-+sqrt+%286%29, since it makes the log of a negative. The final answer is -2+%2B+sqrt+%286%29+

Are you sure you didn't copy this wrong??

R^2 at SCC