# Please solve for x. ln(x+2) - ln(3x + 1) = 4

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ln (x+2) - ln (3x+1) = 4

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

==> ln (x+2)/(3x+1) = 4

==> (x+2)/(3x+1) = e^4

==> x+2 = (3x+1)*e^4

==> x+2 = 3(e^4)x + 3^4

==> 3(e^4)x - x = e^4 - 2

==> x(3e^4 - 1) = e^4 - 2

==> x = (e^4 - 2)/(3e^4 - 1)

ln(x+2)-ln(3x+1) = 4.

Solution:

By property of logarithms ln a- ln b = ln (a/b). And ln a +lnb = ln (a b). So the equation becomes :

ln(x+2) -3ln(3x+1) = 4. Add ln(3x+1)

ln(x+2) = 4+ln(3x+2).

ln(x+2) = lne^4 +ln(3x+1), as ln e^n = n*lne = n.

ln(x+2) = ln(3x+1)e^4.. Take antilogarithms.

x+2 = 3xe^4 +e^4.

x -3xe^4 = e^4-2.

x(1-3e^4) = e^4-2.

x = (e^4-2)/(1-3e^4).

l

First, let's impose the constraints of existence of logarithms:

x+2>0

x>-2

and

3x+1>0

3x>-1

x>-1/3

The common interval of values that satisfies both constraints is (-1/3 , +inf.).

Now, we can solve the equation by subtracting both sides the value ln(x+2).

We'll get:

- ln (3x + 1) = 4 - ln(x+2)

We'll multiply both sides by -1:

ln (3x + 1) = -4 + ln(x+2)

We'll write -4 as:

-4*1 = -4*ln e = ln (e^-4)

We'll re-write the equation:

ln (3x + 1) = ln (e^-4) + ln(x+2)

We'll apply the product property to the right side of the eq.:

ln (3x + 1) = ln [(e^-4)*(x+2)]

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

3x+ 1 = x/e^4 + 2/e^4

We'll subtract both sides x/e^4:

3x - x/e^4 + 1 = 2/e^4

We'll re-write the eq.:

3x*e^4 - x = 2 - e^4

We'll factorize to the left side:

x*(3e^4 - 1) = 2-e^4

We'll divide by (3e^4 - 1):

x = (2-e^4)/ (3e^4 - 1)>-1/3

So, the solution is admissible!

ln(x+2) - ln (3x + 1) = 4

ln([x+2]/[3x+1])=4ln(e)

ln([x+2]/[3x+1])=ln(e^4)

(x+2)/(3x+1)=2.718^4 {because e=2.718281828 }

(x+2)=2.718^4(3x+1)

x+2=163.726x+54.575

162.726x=(-52.575)

x=(-0.323)