1. if $\displaystyle 3^{2x-3}=27$

x=2

2. if $\displaystyle 2^{x-1}= \frac{1}{16}$

x=5

3. if $\displaystyle 2^{x+2}-2^x=12

$

x=-2

4. if $\displaystyle 300(1.05)^n=600$

n=2.78

5. to solve $\displaystyle 3^{x+2}-3^x=216$ you must use the logarithmic method