How do you evaluate #log_(1/6) (1/6)#?
1 Answer
Aug 7, 2016
1
Explanation:
Consider what power the base
#(1/6)# must be raised to, to obtain#1/6# That is
#(1/6)^1=1/6# hence value is 1.In general
#color(red)(|bar(ul(color(white)(a/a)color(black)(log_b(b)=1)color(white)(a/a)|)))#
