Problem: The number of real values of xxx satisfying the equation 22x2−7x+5=12^{2 x^{2}-7 x+5}=122x2−7x+5=1 is:
Answer Choices:
A. 000
B. 111
C. 222
D. 444
E. more than 4 Solution:
22x2−7x+5=1=20∴2x2−7x+5=02^{2 x^{2}-7 x+5}=1=2^{0} \quad \therefore 2 x^{2}-7 x+5=022x2−7x+5=1=20∴2x2−7x+5=0; the solution set of this equation has two members.