How would you solve the equation ln(x-1) - ln(2x+3) = ln(6) - ln(x-2)?

Answer 1

Since the base of all the natural logs are the same, you can get rid of it, leaving:

(x-1) - (2x+3) = (6) - (x-2)
-x - 4 = 8 - x
-4=8
The statement is false, therefore you have no solution.

Answer 2

ln(x - 1) - ln(2x + 3) = ln(6) - ln(x - 2)
same as saying
ln((x - 1)/(2x + 3)) = ln(6/(x - 2))

so now you just need to use

(x - 1)/(2x + 3) = 6/(x - 2)
cross multiply
6(2x + 3) = (x - 1)(x - 2)
12x + 18 = x^2 - 2x - x + 2
12x + 18 = x^2 - 3x + 2
x^2 - 15x - 16 = 0
(x - 16)(x + 1) = 0
x = 16 or -1

since you can't use -1

x = 16

Answer 3

To get rid of the natural log "ln", take all of the coefficients as e^x. For example, ln(x-1) would become e^ln(x-1), and so forth. The e and ln cancel, leaving you with simply x-1. The other logs cancel as well. To finish the equation, solve for x.

Answer 4

ln(x-1/2x+3)=ln(6/x-2)
take 'e' on bothe sides
therefore(x-1/2x+3)=(6/x-2)
solve and get the ans for x

Answer 5

when you subtract logs, it's the log of the quotient, so the left side = ln[(x-1)/(2x+3)], and the right side is ln[6/(x-2)].

So (x-1)/(2x+3) = 6/(x-2)

(x-1)(x-2)=6(2x+3)
x^2-3x+2 = 12x +18
x^2 - 15x -16 = 0

(x-16)(x+1) = 0; x=16 or -1
But you have to check your answers... x=-1 does not work; only x=16 - you can't have ln of -2

Sorry bleh, you had it all the way...

Answer 6

ln(x-1) - ln(2x+3) = ln(6) - ln(x-2)

ln[(x-1)/(2x+3)] = ln(6) - ln(x-2)

ln[(x-1)(x-2)/(2x+3)] = ln(6)

(x-1)(x-2)/(2x+3) = 6

(x^2 -3x +2)/(2x+3) = 6

(x^2 -3x +2) = 12x + 18

x^2 - 15x -16 = 0

Then solve the quadratic so:

(x-16)(x+1) = 0

x=16 and x=(-1)

Answer 7

x = 1/9

Answer 8

um...i havent had to do math in a loong time but...i would say, you have to get rid of those ln's so you can do foil.

sorry not much help.

Answer 9

INXS

Answer 10

i wouldn't.