WE CAN REWRITE THIS EQUATION IN A BASE OF 3

34. (1) SUFFICIENT: We can rewrite this equation in a base of 3:

3

^{b + 2}

= 3

⁵

, which means that b + 2 = 5 and therefore b = 3.

We can plug this value into the equation a = 3

^{b – 1}

to solve for a.

(2) SUFFICIENT: We can set the right side of this equation equal to

the right side of the equation in the quesiton (both sides equal a).

3

^{b – 1}

= 3

^{2b – 4}

, which means that b – 1 = 2b – 4 and therefore b =