Question Number 114561 by Akeyz last updated on 19/Sep/20

Answered by bemath last updated on 19/Sep/20

log _((log _π x))  (log _x π) = −1=log _((log _π x)) (log _π x)^(−1)   log _x π = log _π x   ((ln π)/(ln x)) = ((ln x)/(ln π)) → { ((ln x=ln π⇒x=π)),((ln x=ln ((1/π))⇒x=(1/π))) :}