# ln (x+2) - ln 3x = ln 5

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ln (x+2) - ln 3x = ln 5

We know that :

ln a - ln b = ln a/b

==> ln (x+2)- ln 3x = ln (x+2)/3x = ln 5

==> (x+2)/3x = 5

Now cross multiply:

==> x+2 = 15x

==> 13x = 2

==> x = 2/13

Given:

ln (x+2) - ln 3x = ln 5

We know:

ln a - ln b = ln a/b

Therefore:

We can re write the given equation as:

ln [(x+2)/3x] = ln 5

Taking antilog of both the sides:

(x+2)/3x = 5

==> x + 2 = 5*3x

15x - x = 2

14x = 2

x = 2/14 = 1/7

We'll set the constraints of existence of logarithms:

x+2>0

x>-2

3x>0

x>0

The common interval of admissible values for x is (0,+inf.).

Now,we'll could solve the equation in this way, also:

We'll subtract ln (x+2) both sides:

- ln 3x = ln 5 - ln (x+2)

We'll multiply by -1 both sides:

ln 3x = ln (x+2) - ln 5

We'll apply the quotient property of logarithms:

ln 3x = ln [(x+2)/5]

Because the bases of logarithms are matching, we'll apply the one to one property:

3x = (x+2)/5

We'll cross multiply;

15x = x+2

We'll subtract x both sides:

14x = 2

We'll divide by 14:

x = 2/14

x = 1/7>0

Since the value for x is in the interval of admissible values, the solution is valid.

I assume we have to find x here.

We know that ln a- ln b=ln(a/b)

Therefore ln(x+2)-ln 3x= ln[(x+2)/3x]=ln 5

=>(x+2)/3x=5

=>x+2=15x

=>14x=2

=>x=1/7

**x=1/7**

ln(x+2) -ln3x = ln5. To find x.

Solution:

We know that lnab = lna+lnb , So we can write the equation as>

ln(x+2) - (ln3 + lnx) = ln5.

ln(x+2) -ln3 +lnx = ln5

ln(x+2)+lnx = ln5+ln3

ln[(x+2)x] = ln(5*3) . Take antilogarithms.

(x+2)x = 15

x^2+2x = 15

x^2+2x-15 = 0

(x+5)(x-3) = 0

x+5 = 0 or x-2 = 0

x = -5 or x= 2.