# How do you solve 4(.3)^x = 1.2^(x+2)?

Jun 21, 2016

$\frac{\ln 5 - \ln 3}{\ln 2}$

#### Explanation:

take logs

$\ln \left(4 {\left(0.3\right)}^{x}\right) = \ln \left({1.2}^{x + 2}\right)$
$\ln \left(4\right) + \ln \left({0.3}^{x}\right) = \ln \left({1.2}^{x + 2}\right)$
$\ln \left(4\right) + x \ln \left(0.3\right) = \left(x + 2\right) \ln \left(1.2\right)$
$x \left(\ln \left(0.3\right) - \ln \left(1.2\right)\right) = 2 \ln \left(1.2\right) - \ln \left(4\right)$
$x \ln \left(\frac{0.3}{1.2}\right) = \ln \left({1.2}^{2} / 4\right)$
$x \ln \left(\frac{1}{4}\right) = \ln \left(\frac{9}{25}\right)$
$x = \ln \frac{\frac{9}{25}}{\ln} \left(\frac{1}{4}\right) = \frac{\ln 9 - \ln 25}{\ln 1 - \ln 4}$
$= \frac{2 \ln 3 - 2 \ln 5}{0 - 2 \ln 2}$
$= \frac{\ln 5 - \ln 3}{\ln 2}$

Jun 21, 2016

$x = 0.736966$

#### Explanation:

$4 {\left(.3\right)}^{x} = {1.2}^{x + 2} = {\left(4 \times .3\right)}^{x + 2} = {4}^{x + 2} \times {\left(.3\right)}^{x + 2}$

then

$4 {\left(.3\right)}^{x} = {4}^{x + 2} \times {\left(.3\right)}^{x + 2} = {4}^{2} \times {\left(0.3\right)}^{2} \times {4}^{x} \times {\left(.3\right)}^{x}$

elliminating ${\left(.3\right)}^{x}$ in both sides

$4 = {4}^{2} \times {\left(0.3\right)}^{2} \times {4}^{x} \to {4}^{x} = \frac{1}{4 \times {\left(0.3\right)}^{2}}$

and finally

$x = - {\log}_{e} \frac{4 \times {\left(0.3\right)}^{2}}{\log} _ e 4 = 0.736966$