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

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

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Answer 1 · 2015-12-01T10:01:05

$x = 4 + \frac{40 ln 2}{ln 3 - 6 ln 2}$ $x= 4 + (40ln2)/(ln3-6ln2)$

Explanation:

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

Now taking natural log on both sides, it would be $(6 x - 16) ln 2 = (- x + 4) ln 3$ $(6x-16) ln 2= (-x+4)ln3$

$\frac{6 x - 16}{- x + 4} = \frac{ln 3}{ln 2}$ $(6x-16)/(-x+4)= ln3 /ln2$ , Or, $- \frac{40}{- x + 4} = \frac{ln 3}{ln 2} - 6 = \frac{ln 3 - 6 ln 2}{ln 2}$ $-40/(-x+4)= ln3/ln2 -6= (ln3-6ln2)/ln2$

$- x + 4 = \frac{- 40 ln 2}{ln 3 - 6 ln 2}$ $-x+4= (-40ln2)/(ln3-6ln2)$

$x = 4 + \frac{40 ln 2}{ln 3 - 6 ln 2}$ $x= 4 + (40ln2)/(ln3-6ln2)$

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

1 Answer

Explanation:

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