# How do you solve 2^ { x } \cdot 3^ { 2x - 1} = 5^ { x + 1}?

Jul 5, 2017

$x = = 2.1141$

#### Explanation:

Taking logarithm (base10) on both sides of ${2}^{x} \cdot {3}^{2 x - 1} = {5}^{x + 1}$, we get

$x \log 2 + \left(2 x - 1\right) \log 3 = \left(x + 1\right) \log 5$

or $x \left(\log 2 + 2 \log 3 - \log 5\right) = \log 5 + \log 3$

or $x \log \left(\frac{2 \cdot {3}^{2}}{5}\right) = \log 15$

or $x \log 3.6 = \log 15$

or $x = \log \frac{15}{\log} 3.6 = \frac{1.1761}{0.5563} = 2.1141$