We'll change the base 4 of the logarithm into the base 2:

log[base 4](x-1) = log[base 2](x-3)

log 2 (x-1) = log 4 (x-1)*log 2 (4)

But log 2 (4) = log 2 (2^2)

We'll apply the power property of logarithms and we'll get:

log 2 (2^2) = 2*log 2 (2)

But log 2 (2) = 1

log 2 (2^2) = 2

log[base 4](x-1) = log[base 2](x-1)/2

The equation will become:

{log[base 2](x-1)}/2 = log[base 2](x-3)

log[base 2](x-1) = 2*log[base 2](x-3)

log[base 2](x-1) = log[base 2][(x-3)^2]

We'll apply one to one property:

x-1 = (x-3)^2

We'll expand the square:

x-1 = x^2 - 6x + 9

We'llmove all terms to one side:

x^2 - 7x + 10 = 0

We'll apply the quadratic formula:

x1 = [7+sqrt(49-40)]/2

x1 = (7+3)/2

**x1 = 5**

x2 = (7-3)/2

**x2 = 2**

Now, we'll check if the found solutions are convenient and respect the constraints of existence of logarithms above:

The constraints of existence of logarithms are:

x-1>0

x>1

x-3>0

x>3

So, all solutions have to be more than the value 3.

**Because the second solution, x2=2<3, is not acceptable, so the only solution of the equation is x=5.**