How do you solve #16^(d-4)=3^(3-d)#?

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Answer 1 · 2016-10-19T10:42:40

#d=3.7162#

As #16^(d-4)=3^(3-d)#, taking log to the base #10# on both sides, we get

#(d-4)log16=(3-d)log3#

or #d xxlog16-4log16=3 xx log3-d xxlog3#

or #d(log16+log3)=3xxlog3+4log16#

or #d=(3xxlog3+4log16)/log48#

= #(3xx0.4771+4xx1.2041)/1.6812#

= #(1.4313+4.8164)/1.6812#

= #6.2477/1.6812=3.7162#