# How to determine all real values of a if the equation 2lg(x-1)=lg(ax-3) has a single real solution?

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The equation 2*lg(x-1) = lg (ax-3) has two equal roots.

2*lg(x-1) = lg (ax-3)

=> lg ( x-1)^2 = lg (ax - 3)

taking the antilog of both the sides

=> (x - 1)^2 = ax - 3

=> x^2 - 2x + 1 = ax - 3

=> x^2 - x(2 + a) + 4 = 0

As this quadratic equation has two equal and real roots

(2 + a)^2 - 4*1*4 = 0

=> 4 + a^2 + 4a - 16 = 0

=> a^2 + 4a - 12 = 0

=> a^2 + 6a - 2a - 12 = 0

=> a(a + 6) - 2( a + 6) = 0

=> (a - 2)(a + 6) = 0

=> a = 2 and a = -6

**The required values of a are 2 and -6.**

Before finding the value of "a", we'll impose the constraints of existence of logarithms:

x-1>0

x>1

ax-3>0

x>3/a

Now, we'll solve the equation. For the beginning, we'll apply the power rule of logarithms:

2lg(x-1)=lg(ax-3)

lg(x-1)^2 = lg(ax-3)

Since the bases are matching, we'll apply one to one property of logarithms:

(x-1)^2 = ax - 3

We'll expand the square:

x^2 - 2x + 1 = ax - 3

We'll move all terms to one side:

x^2 - x(2 + a) + 4 = 0

This equation has a single real root if delta is equal to zero:

delta = (2+a)^2 - 16

We'll expand the square:

4 + 4a + a^2 - 16 = 0

a^2 + 4a - 12 = 0

a1 = [-4+sqrt(16 + 48)]/2

a1 = (-4+8)/2

a1 = 2

a2 = -12/2

a2 = -6

If a = 2, the constraints of existence of logarithms are: x>1 and x>3/2, so the interval of admissible solutions is (3/2 ; +infinite).

If a = -6

x>1

x>-1/2

The interval of admissible solutions is (1 ; +infinite).

**The values of a are: a = 2 and a = -6.**