# How do you solve 4^(x + 7) = 6^(x – 1)?

Apr 10, 2016

$\text{ } x = \frac{8 \ln \left(4\right)}{\ln \left(6\right) - \ln \left(4\right)} + 1$

$\text{ } x \approx 28.352$ to 3 decimal places

#### Explanation:

Take logs

$\text{ } \ln \left({4}^{x + 7}\right) = \ln \left({6}^{x - 1}\right)$

$\text{ } \implies \left(x + 7\right) \ln \left(4\right) = \left(x - 1\right) \ln \left(6\right)$

$\text{ } \implies \frac{x + 7}{x - 1} = \ln \frac{6}{\ln} \left(4\right)$.............................(1)

But $x - 7$ can be written as $x - 1 + 8$

Write equation (1) as

$\text{ } \implies \frac{x - 1 + 8}{x - 1} = \ln \frac{6}{\ln} \left(4\right)$

$\text{ } \implies \frac{x - 1}{x - 1} + \frac{8}{x - 1} = \ln \frac{6}{\ln} \left(4\right)$

$\text{ } 1 + \frac{8}{x - 1} = \ln \frac{6}{\ln} \left(4\right)$

$\text{ } \frac{8}{x - 1} = \ln \frac{6}{\ln} \left(4\right) - 1$

$\text{ } \frac{8}{x - 1} = \frac{\ln \left(6\right) - \ln \left(4\right)}{\ln} \left(4\right)$

$\text{ } x - 1 = \frac{8 \ln \left(4\right)}{\ln \left(6\right) - \ln \left(4\right)}$

$\text{ } x = \frac{8 \ln \left(4\right)}{\ln \left(6\right) - \ln \left(4\right)} + 1$

$\text{ } x \approx 28.352$ to 3 decimal places