# How do you solve 2.4^(3x+1)= 9?

Sep 12, 2015

$x = \frac{\log 9 - \log 8}{6 \log 2} = 0 , 02832$

#### Explanation:

Using laws of exponents and indices you may write the equation as
${2}^{1} \cdot {2}^{6 x} \cdot {2}^{2} = {3}^{2}$
$\therefore {2}^{6 x + 3} = {3}^{2}$
$\therefore {2}^{6 x} = \frac{{3}^{2}}{{2}^{3}} = \frac{9}{8}$
Now taking the logarithm on both sides and sing laws of logs we get
$6 x \log 2 = \log 9 - \log 8$
$\therefore x = \frac{\log 9 - \log 8}{6 \log 2} = 0 , 02832$