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

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.