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log₃ (4*3^{x- 1}  - 1)   = 2x - 1   write this in exponential form

3^(2x - 1)  =  4*3^(x - 1) - 1

3^(2x) / 3  =  4*3^x / 3  - 3/3       multiply through by 3

3^(2x)  = 4*3^x  -  3

(3^2)^x  = 4*3^x - 3

(3^x)^2  = 4*(3^x)  -  3

Let  a  =  3^x

a^2  =  4a  - 3

a^2  - 4a + 3   =  0       factor

(a - 3)  (a - 1)  =  0

Set  each factor to 0  and solve for a and we get that

a  =  3    or    a  =1        which means that

1 =  3^x    ⇒  x = 0      or  that

3  =  3^x   ⇒   x  =  1

Both solutions solve the original equation

I'm sorry...but I don'tknow what "domain of the equation"  means....unless we require that 

4*3^(x - 1)  -  1  >  0    if so.....then

4*3^(x - 1)  .  1

3^(x - 1)  > 1/4     take the log of both sides

(x - 1) log(3)  > log (1/4)

log (3) *  (x)  >  log (1/4) + log (3)

x > [ log (1/4) + log (3) ] / log (3)  

x >  log (3/4) / log(3)

x > ≈  -0.262  ⇒    this is the domain