Let #f(x) = klog_2x# Given that #f^-1 (1) = 8# , what is the value of k?

1 Answer
Jan 26, 2016

#k=1/3#

Explanation:

Given #f(x)=klog_2x# and #f^-1(1)=8#

We know that, if #f^-1(x)=y# then #f(y)=x#.
So, in the second equation, this means that #f(8)=1#
We have the first equation there, so we substitute #x=8# and #f(x)=1# to get
#1=klog_2(8)#

I'm sure you know what to do from here to get the above answer.

Hint:- #log_xy^z=zlog_xy#
#log_x(x)=1#