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 $d=(3xxlog3+4log16)/log48$

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

= $(1.4313+4.8164)/1.6812$

= $6.2477/1.6812=3.7162$

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