You need to solve for x the given equation, such that:

`log(10x + 4) = log(4x - 2) => 10x + 4 = 4x - 2`

`10x - 4x = -4 - 2 => 6x = -6 => x = -1`

You need to test the value `x = -1` in equation, such that:

`log(10*(-1) + 4) = log(4*(-1) - 2) => log -6 = log -6`

You should notice that the value `log- 6` is invalid, hence, `x = -1` is not accepted as solution.

**Hence, evaluating the solution to the given equation yields that there are no solutions.**

First, we'll have to impose constraints of existance of logarithms.

The first constraint: 10x+4>0

We'll divide by 2 both sides:

5x+2>0

5x>-2

We'll divide by 5 both sides:

x>-2/5

The second constraint: 4x-2>0

We'll divide by 2 boh sides:

2x - 1>0

We'll add 1 both sides:

2x>1

We'll divide by 2 both sides:

x>1/2

The values of x which satisfy both constraints belong to the interval (1/2 , +inf.)

Since the bases of logarithms are matching, we'll solve the equation, using the one to one property of logarithms:

10x + 4 = 4x - 2

We'll factorize by 2:

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

We'll divide by 2 both sides:

5x+2 = 2x-1

We'll subtract 2x both sides:

5x - 2x + 2 = -1

3x + 2 = -1

We'll subtract 2 both sides:

3x = -3

We'll divide by 3:

x = -1 < 1/2

The x value doesn't belong to the interval of admissible values, so, the equation has no solution