SOLUTION: ln(x+2)=ln(2x-1)+3

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Question 29425: ln(x+2)=ln(2x-1)+3
Found 2 solutions by sdmmadam@yahoo.com, longjonsilver:
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
ln(x+2)=ln(2x-1)+3
ln(x+2)-ln(2x-1) = 3----(1) (here the base is 10 )
log[(x+2)/(2x-1)] =3 (using log(a)-log(b) =log(a/b) )
(x+2)/(2x-1)=10^3
(x+2) = (2x-1)(10^3)
x+2 = 2X(10)^3(x)-(10^3)
x+2 = 2000x-1000
2+1000=2000x-x
1002=1999x
Therefore x = 1002/1999
Answer: x = 1002/1999
Verification: Putting x = 1002/1999 in (1)
LHS = ln(x+2)-ln(2x-1)
=log[(1002/1999)+2] - log[(2004/1999)-1]
=log{[1002+(2X1999)]/1999}- log{[(2004-1999)]/1999}
=log{[1002+3998]/1999}- log{(5)/1999}
=log{(5000)/1999}- log{(5)/1999}
=(log5000-log1999)-(log5-log1999)
=log5000-log1999-log5+log1999
=log5000-log5
=log(5000/5)
=log(1000)
=log[(10)^3]
=3Xlog(10)
=3X1 (since log10 to the same base 10 is 1)
=3 = RHS

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
ln(x+2)=ln(2x-1)+3
ln(x+2) - ln(2x-1) = 3
+ln%28%28x%2B2%29%2F%282x-1%29%29+=+3+
+%28x%2B2%29%2F%282x-1%29+=+e%5E3+

Now, e%5E3 is just a number...JUST a number!

So,
+%28x%2B2%29+=+e%5E3%282x-1%29+
+x%2B2+=+2e%5E3x-e%5E3+
+x%2B2-2e%5E3x+=+-e%5E3+
+x-2e%5E3x+=+-2-e%5E3+
+x%281-2e%5E3%29+=+-2-e%5E3+
+x+=+%28-2-e%5E3%29%2F%281-2e%5E3%29+
+x+=+%28-%282%2Be%5E3%29%29%2F%28-%282e%5E3-1%29%29+
+x+=+%282%2Be%5E3%29%2F%282e%5E3-1%29+

so either leave this as it is...it is the PRECISE answer. If you use a calulator, you will have to round up the answer, since it is a never-ending number... x = 0.563822605 to 9dp.

PS the other solution by sdmmadam is wrong because on line 3 the "ln" becomes "log" which is wrong. The actual mechanics is correct though...just that the x value is wrong.

jon.