# Solve the equation 2lnx-ln(x+2)=0 ?

### 5 Answers

- PinkgreenLv 74 months ago
The given equation

=>

ln[(x^2)/(x+2)]=0

=>

(x^2)/(x+2)=e^0=1

=>

x^2-x-2=0

=>

(x-2)(x+1)=0

=>

x=2 or x=-1(rejected)

x=2 is the real solution.

- rotchmLv 74 months ago
Rewrite as

2 ln(x) = ln(x+2)

ln(x²) = ln(x+2). If ln(A) = ln(B) then A=B, thus,

x² = x+2. This is a quadratic,ich you know how to solve. Verify if your solutions belong to the domain of the original eqs. Done!

- Engr. RonaldLv 74 months ago
2ln(x)-ln(x+2)=0

ln(x)^2 - ln(x + 2) = 0

ln[x^2/(x + 2)] = e^0

x^2/(x + 2) = 1

x^2 = x + 2

x^2 - x - 2 = 0

(x + 1)(x - 2) = 0

x = - 1, x = 2

discard x = - 1 since it's a negative value.

Answer is x = 2

- Wayne DeguManLv 74 months ago
2lnx => ln(x²)

so, ln(x²) - ln(x + 2) = 0

Then, ln[x²/(x + 2)] = 0

=> x²/(x + 2) = 1

i.e. x² = x + 2

so, x² - x - 2 = 0

or, (x - 2)(x + 1) = 0

Then, x = 2 or x = -1...not possible as x > 0

Hence, x = 2 is the only solution

:)>

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- 4 months ago
a * ln(b) = ln(b^a)

ln(a) - ln(b) = ln(a/b)

ln(a) = b, then e^b = a

2 * ln(x) - ln(x + 2) = 0

ln(x^2) - ln(x + 2) = 0

ln(x^2 / (x + 2)) = 0

x^2 / (x + 2) = e^0

x^2 / (x + 2) = 1

x^2 = x + 2

x^2 - x - 2 = 0

x = (1 +/- sqrt(1 + 8)) / 2

x = (1 +/- 3) / 2

x = -2/2 , 4/2

x = -1 , 2

x = -1 doesn't work in the original problem, unless you're permitted to analyze complex logarithms (which I doubt)

x = 2.

Test it out to make sure it works

2 * ln(2) - ln(2 + 2) = 0

2 * ln(2) - ln(4) = 0

2 * ln(2) - ln(2^2) = 0

2 * ln(2) - 2 * ln(2) = 0

0 = 0

It works

2 * ln(-1) - ln(-1 + 2) = 0

2 * ln(-1) - ln(1) = 0

2 * ln(-1) - 0 = 0

2 * ln(-1) = 0

ln(-1) = 0

Now, unless we're dealing with complex numbers, we can't go any further. ln(t) is only defined when t > 0. In this case, if we decided to move forward, we'd have e^(-1) = 0, which isn't true. Had I done steps differently, however, we could have had a false answer

2 * ln(-1) - ln(-1 + 2) = 0

ln((-1)^2) - ln(1) = 0

ln(1) - ln(1) = 0

0 = 0

See what I did? I took something that wasn't defined, 2 * ln(-1) and I screwed around with it until it became ln((-1)^2), which is just ln(1). I changed something that was fundamental and if you're not careful, you can easily find yourself doing that and making infuriating mistakes. We could explore the matter more deeply with complex numbers, but that's not necessary. x = 2 is the only valid answer here.