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

Dec 1, 2015

$x = 4 + \frac{40 \ln 2}{\ln 3 - 6 \ln 2}$

#### Explanation:

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

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

$\frac{6 x - 16}{- x + 4} = \ln \frac{3}{\ln} 2$, Or, $- \frac{40}{- x + 4} = \ln \frac{3}{\ln} 2 - 6 = \frac{\ln 3 - 6 \ln 2}{\ln} 2$

$- x + 4 = \frac{- 40 \ln 2}{\ln 3 - 6 \ln 2}$

$x = 4 + \frac{40 \ln 2}{\ln 3 - 6 \ln 2}$