Problem: If logb2βx+logx2βb=1,b>0,bξ =1,xξ =1, then x equals:
Answer Choices:
A. 1/b2
B. 1/b
C. b2
D. b
E. bβ
Solution:
Let logbβx=m and let logxβ2b=n. Then x=b2m and b=x2n.
β΄(x2n)2 m=b2 mβ΄x4mn=xβ΄4mn=1β΄n=4 m1β
β΄m+4m1β=1β΄m=21ββ΄x=b