Solve this logarithm : 2ln(x) - ln(x+1) = 0?

Answer 1

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

lnx^2 - ln(x+1) = 0

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

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

x^2 = x+1

x^2 - x - 1 = 0

a = 1 b= -1 c= -1

x = [-(-1) +- sqrt(1-4(1)(-1)]/2(1)

x = 1/2 +- (1/2)sqrt(5)

Answer 2

Combine terms using the fact that the difference of logs is equal to the log of the quotient of the arguements, and a coeffeicnt on a log is equal to an exponent on the arguement. So 2 lnx = lnx^2
and ln (x^2/(x+1)) = 0
Since e^ln(u) = u for all u in the domain, raise each side to e power and get
x^2 / (x+1) = 1 (note: e^0 = 1, since N^0 = 1 for all nonzero N)
Multiply each side by the denominator and you get
x^2 = x + 1
Write it in standard form
x^2 - x - 1 = 0 and solve. (This, btw, looks like the equation which describes the golden ratio) Since it doesn't factor, use the quadratic formula, as others have shown, and be sure to throw out any negative values of x, as extraneous roots, since ln is defined only for positive x.

Answer 3

2ln(x) - ln(x+1) = 0
2ln(x) = ln(x+1)
ln(x²) = ln(x+1)
x² = x + 1
x² - x - 1 = 0
x = (1 ± √(1 + 4))/2
x = (1 + √5)/2

Only the positive answer is valid since you can only have a log of a positive number.

Answer 4

2ln(x) = ln(x^2), so...

ln(x^2) = ln(x+1)

x^2 = x + 1.

Now solve the quadratic. (Note to poster above me - you factored this wrong. I think it should be (1 + sqrt(5))/2.)

Answer 5

Remake from identities
ln[x^2] = ln[x+1]
raise to e on both sides
x^2 = x+1
x^2 -x -1 = 0
x = {--1+-sqr[(-1)^2 -4(1)(1)]}/2(1)
= {1 +- sqr[1+4]}/2 = {1 +-sqr[5]}/2

Answer 6

ln((x^2) = ln(x+1)
Taking the antilog: x^2 = x + 1
x^2 -x - 1 = 0

Solving using the quadratic equation:
a = 1
b = -1
c = -1
b^2 - 4ac = 1 + 4 = 5
The full equation:
[-(-1) +/- sqrt 5]/-1
-1 +/- sqrt 5 = -1 +2.24 or -1 =-2.24
x = 1.24 or -3.24

Answer 7

2ln(x) - ln(x+1) = 0
ln(x^2/(x+1)=0
x^2/(x+1)=1
x^2=x+1
x^2-x-1=0
x=(1+/-√(1+4))/2
x=.5+/-.5√5
x=.5+.5√5, .5-.5√5

Answer 8

Swimmer is correct till x^2-x-1=0, but you can not factorize that with (x+1)(x-1)=0 then using X*=(-b+-sqrt(b^2-4ac))/2a yo have
x*=(-(-1)+-sqrt((-1)^2-4*1*(-1))/(2(1))
x*=(sqrt(5)+1)/2 or X*=(-sqrt(5)+1)/2

Answer 9

2lnx = ln(x+1)
lnx² = ln(x+1)
x² = x+1

By definition of phi, the Golden Ratio, the solutions are
x=phi
x=(1-phi)

Answer 10

0-.02=3=11^