# solve for x 2 log x - log ( x + 1) = log 4 - log 3please I need details

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The equation 2 log x - log ( x + 1) = log 4 - log 3 has to be solved for x.

Use the property of logarithm a*log b = log b^a and log a - log b = log(a/b)

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

log x^2 - log(x + 1) = log (4/3)

log(x^2/(x + 1)) = log (4/3)

This gives:

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

3x^2 = 4x + 4

3x^2 - 4x - 4 = 0

3x^2 - 6x + 2x - 4 = 0

3x(x - 2) + 2(x - 2) = 0

(3x + 2)(x - 2) = 0

x = -2/3 and x = 2

As the logarithm is not defined for a negative number ignore the root x = -2/3.

The solution of 2 log x - log ( x + 1) = log 4 - log 3 is x = 2

We'll use the power property of logarithms for the 1st term:

a*log b = log (b^a)

2*log x = log (x^2)

Since the bases of logarithms of the left side are matching, we'll transform the difference into a product:

log a - log b = log (a/b)

log (x^2) - log (x+1) = log [x^2/(x+1)]

Since the bases of logarithms of the right side are matching, we'll transform the difference into a product:

log 4 - log 3 = log (4/3)

We'll re-write the equation:

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

Since the bases of logarithms are matching, we'll apply one to one rule:

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

We'll cros multiply and we'll get:

3x^2 = 4sx + 4

We'll move all terms to the left side:

3x^2 - 4x - 4 = 0

We'll apply quadratic formula:

x1 = [4 + sqrt(16+48)]/6

x1 = (4+8)/6

x1 = 2

x2 = (4-8)/6

x2 = -2/3

Since the values of the solutions must be positive, for the given logarithmic functions to exist, we'll reject the negative value for x.

**The only solution for the given equation is x = 2.**