# What is x if lg(10x+4)=lg(4x-2) ?

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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.**

In this question we have to solve for x, given lg(10x+4) = lg(4x-2)

Now the logarithm is defined only for a positive number so we have the constraint that 10x + 4 > 0 => x > -4 /10 and 4x-2 >0 => x > 2/4 . So we consider values only which are greater than 2/4

Now lg(10x+4) = lg(4x-2)

take the anti-log of both the sides

=> 10x + 4 = 4x - 2

bring the terms with x to one side and the numeric terms to the other side

=> 10x - 4x = -2 -4

=> 6x = -6

=> x = -1

But we can only consider value of x greater than 1/2.

**Therefore the equation has no solutions.**