**Question 1**

Solve for the unknown variable:

- $3^{x-1}-27=0$
- $27(4^{x})=(64)3^{x}$
- $\sqrt{2x-5}=2-x$

**Question 2**

- Show that $\sqrt{\dfrac{3^{x+1}-3^{x}}{3^{x-1}}+3}$ is equal to $\text{3}$
- Hence solve for x$\sqrt{\dfrac{3^{x+1}-3^{x}}{3^{x-1}}+3}=\left(\dfrac{1}{3}\right)^{x-2}$