Solve: log 2 (3x ) - log 4 ( x-2) = 3

First, we'll impose constraints of existence of
logarithms:

3x>0

x>0

x-2>0

x>2

The
common interval of admissible values is (2, +inf.)

Now,
we'll solve the equation.

log 2  (3x)  - log 4 (x-2) =
3

We'll add log 4 ( x-2) both
sides:

log 2  (3x) = log 4 (x-2) +
3

We'll write 3 = 3*1 = 3*log 4 4 = log 4
4^3

We'll re-write the
equation:

log 2  (3x) = log 4 (x-2) + log 4
4^3

We'll use the product property of
logarithms:

log 2  (3x) = log 4
[(x-2)*4^3]

We'll change the base 2 of the logarithm from
the left side, into the base 4:

log 4 (3x) = log 2  (3x) *
log 4 2

But log 4 2 = 1/log 2 4 = 1/log 2 2^2 =
1/2

2*log 4 (3x) = log 2 
(3x)

Now, we'll re-write the equation with all logarithms
having matching bases:

2*log 4 (3x) = log 4
[(x-2)*4^3]

log 4 (9x^2) = log 4
[(x-2)*4^3]

We'll use the one to one
property:

9x^2 = 64x -
128

We'll move all terms to one
side:

9x^2 - 64x + 128 = 0

x1
= (64+sqrt512)/18

x1 =
(64+16sqrt2)/18

x1 =
16(4+sqrt2)/18

x1 =
(32+8sqrt2)/9

x2 =
(32-8sqrt2)/9

Since both
solutions are in the interval (2,+inf.), they are
valid.