How do you solve for k in #(1/2)^(10/k) = 0.8#?

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Answer 1 · 2016-05-26T10:10:42

#k=31.06#

#(1/2)^(10/k)=0.8=4/5#. Now taking log on both sides

#(10/k)log(1/2)=log(4/5)# or

#(10/k)=log(4/5)/log(1/2)# or

#10=k*log(4/5)/log(1/2)# or

#k=10/(log(4/5)/log(1/2))# or

#k=10*log(1/2)/log(4/5)=10*(-0.3010)/(-0.09691)#

or #k=31.06#