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)
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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)
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.
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
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