Problem: The number of real values of x satisfying the equality (logaβx)(logbβx)=logaβb, where a>0,b>0,aξ =1,bξ =1, is:
Answer Choices:
A. 0
B. 1
C. 2
D. a finite integer greater than 2
E. not finite
Solution:
Let logaβx=yβ΄x=ayβ΄logbβx=logbβay=ylogbβa
β΄{(logaβx)(logbβx)=logaβb} implies {yβ
ylogbβa=logaβb}
Since logbβa=1/logaβb,y2=(logaβb)2,y=logaβb or y=βlogaβb
β΄logaβx=logaβb or logaβx=logaβbβ1β΄x=b or x=bβ1=1/b
or
Let logbβx=yβ΄x=byβ΄(logaβby)(logbβby)=logaβb
(ylogaβb)(ylogbβb)=logaβb,y2logaβb=logaβbβ΄y2=1β΄y=1 or y=β1β΄x=b or x=b^
(logbβx)(logaβx)=logaβx(logbβx)=logaβbβ΄xlogbβx=b
β΄(logbβx)(logbβx)=logbβb=1β΄logbβx=1 or logbβx=β1β΄x=b or x=b^