Problem:
How many real numbers x satisfy the equation 32x+2−3x+3−3x+3=0?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Since
0=32x+2−3x+3−3x+3=32(3x)z−28(3x)+3=9(3x)2−28(3x)+3=(3x−3)(9(3x)−1)
we must have 3x=3 or 3x=91. Thus, x=1 or −2 are the only solutions.