How do you solve #9^(5k)+8=50#?

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Answer 1 · 2016-10-11T09:22:36

#k=0.3402#

As #9^(5k)+8=50#,

#9^(5k)=50-8=42#

or taking logarithm to base #10#, we get

#5klog9=log42#

or #k=log42/(5log9)#

or #k=1.62325/(5xx0.95424)#

or #k=0.3402#