How do you solve $3·8^(2x-5) = 2·3^(-x+5)$ ?

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Answer 1 · 2015-12-01T10:01:05

$x= 4 + (40ln2)/(ln3-6ln2)$

The given equation can be written as $3. 2^(6x-15) =2.3^(-x+5)$
Or, $2^(6x-16) = 3^(-x+4)$ on dividing both sides by 6.

Now taking natural log on both sides, it would be $(6x-16) ln 2= (-x+4)ln3$

$(6x-16)/(-x+4)= ln3 /ln2$ , Or, $-40/(-x+4)= ln3/ln2 -6= (ln3-6ln2)/ln2$

$-x+4= (-40ln2)/(ln3-6ln2)$

$x= 4 + (40ln2)/(ln3-6ln2)$

How do you solve 3·8^(2x-5) = 2·3^(-x+5)3·8^(2x-5) = 2·3^(-x+5)?