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

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.