# How do you solve 4(7^(x + 2)) = 9^(2x - 3)?

Apr 22, 2018

$x = \frac{- 3 \ln \left(9\right) - 2 \ln \left(7\right) - \ln \left(4\right)}{\ln \left(7\right) - 2 \ln \left(9\right)}$

#### Explanation:

you have to log the equations

$4 \cdot {7}^{x + 2} = {9}^{2 x - 3}$

Use either natural logs or normal logs $\ln$ or $\log$ and log both sides

$\ln \left(4 \cdot {7}^{x + 2}\right) = \ln \left({9}^{2 x - 3}\right)$

First use the log rule that states $\log a \cdot b = \log a + \log b$

$\ln \left(4\right) + \ln \left({7}^{x + 2}\right) = \ln \left({9}^{2 x - 3}\right)$

Remember the log rule that states $\log {x}^{4} = 4 \log x$

$\ln \left(4\right) + \left(x + 2\right) \ln \left(7\right) = \left(2 x - 3\right) \ln \left(9\right)$

$\ln \left(4\right) + x \ln \left(7\right) + 2 \ln \left(7\right) = 2 x \ln \left(9\right) - 3 \ln \left(9\right)$

Bring all the $x \ln$ terms to one side

$x \ln \left(7\right) - 2 x \ln \left(9\right) = - 3 \ln \left(9\right) - 2 \ln \left(7\right) - \ln \left(4\right)$

Factorise the x out

$x \left(\ln \left(7\right) - 2 \ln \left(9\right)\right) = \left(- 3 \ln \left(9\right) - 2 \ln \left(7\right) - \ln \left(4\right)\right)$

$x = \frac{- 3 \ln \left(9\right) - 2 \ln \left(7\right) - \ln \left(4\right)}{\ln \left(7\right) - 2 \ln \left(9\right)}$

Solve on the calculator using the ln button or if your calculator doesn't have it use the log base 10 button.