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

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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.