Write the equation

We need to solve $\log_4(x-1)=\log_2(x-3).$

Apply domain conditions

For the logarithms to be defined: $x-1>0 \quad \text{and} \quad x-3>0.$ So, $x>1 \quad \text{and} \quad x>3 \implies x>3.$

Convert all logarithms to the same base

Since $4=2^2$, $\log_4(x-1)=\frac{\log_2(x-1)}{\log_2 4}=\frac{\log_2(x-1)}{2}.$

Therefore the equation becomes $\frac{1}{2}\log_2(x-1)=\log_2(x-3).$

Multiply by 2

$\log_2(x-1)=2\log_2(x-3).$

Using $2\log_2(x-3)=\log_2\big((x-3)^2\big)$, $\log_2(x-1)=\log_2\big((x-3)^2\big).$

Equate arguments

Since logarithm with the same base is one-one on its domain, $x-1=(x-3)^2.$

Solve the quadratic equation

$x-1=x^2-6x+9$ $0=x^2-7x+10$ $x^2-7x+10=0$ $(x-5)(x-2)=0.$

So, $x=5 \quad \text{or} \quad x=2.$

Check domain

Domain requires $x>3$.

• $x=5$ is valid.
• $x=2$ is invalid.

Hence only one solution exists.

Conclusion

The number of solutions is $\boxed{1}.$