How do I solve this logarithm?

5

log(x-4) - log(x+2)=3
log((x-4)(x+3)) = 3
10^3 = x^2 -x - 12
x^2 - x - 1012 = 0
x = -31.3158
x = 32.3158765

x = 32.3158765

log [ (x - 4) / (x + 2) ] = 3
(x - 4) / (x + 2) = 1000
x - 4 = 1000x + 2000
- 2044 = 999 x
x = - 2044 / 999

log(x-4)-log(x+2)=3
or, log (x-4)/(x+2)=3
or, (x-4) / (x+2)=10^3
or, x-4= 1000x +2000
therefore x = -2.006

This problem has no solution because we cannot accept the negative value of x in this problem.

log(x - 4) - log(x + 2) = 3
log(x - 4)/(x + 2) = 3
(x - 4)/(x + 2) = 10^3
x - 4 = 1000(x + 2)
x - 4 = 1000x + 2000
x - 1000x = 4 + 2000
-999x = 2004
x = 2004/-999
x = -668/333

log(x-4) - log(x+2) = 3
log((x-4)/(x+2)) =3
(x-4)/(x+2) = 10^3
x - 4 = 1000x + 2000
-999x =2004
x = -2.006

Letting b = base of the logarithm (for the natural log, it
would be "e"), logarithms have the following properties:

1. log(xy) = log(x) + log(y)
2. log(x/y) = log(x) – log(y)
3. b^[log(x)] = x
4. log(b^x) = x
5. log(x^y) = y[log(x)]

Here we shall leverage (2) and (3) above to help
solve your equation:

log(x – 4) – log(x + 2) = 3
log[(x – 4)/(x + 2)] = 3

b^(log[(x – 4)/(x + 2)]) = b^3

(x – 4)/(x + 2) = b^3
x – 4 = (b^3)(x + 2)
x – 4 = (b^3)x + 2(b^3)
x – (b^3)x = 2(b^3) + 4
x[1 – (b^3)] = 2(b^3) + 4

x = [2(b^3) + 4]/[1 – (b^3)]

If our base b = 10, then b^3 = 1000, so that
we have

x = [2000 + 4]/[1 – 1000]
= 2004/(–999)
= –668/333

x = –668/333

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log[(x-4)/(x+2)]=3
(x-4)(x+2)=1000
x^2-2x-8=1000
x^2-2x-1008=0
x~32.8 or x~-30.76
log(-ve) not defined
so x~32.8

==>log(x-4)-log(x+2)=3
==>log((x-4)/(x+2))=3
==>10^3=(x-4)(x-2)
==>1000=x^2-6x+8
==>x^2-6x-992=0
==>now factorize and you will have the answer

I think to get rid of the logs you have to use ln or maybe to get rid of the ln you have to do E^(x-4)- E^x+2)