# How do you solve 3.5^(x+2) = 1.75^(x+3)?

Oct 5, 2015

$x = \frac{\ln \left(\frac{7}{16}\right)}{\ln} \left(2\right) \approx - 1.19265$

#### Explanation:

Take the natural logarithm (or any logarithm) of both sides and use properties of logarithms to help you solve for $x$.

$\ln \left({3.5}^{x + 2}\right) = \ln \left({1.75}^{x + 3}\right) \setminus \rightarrow \left(x + 2\right) \ln \left(3.5\right) = \left(x + 3\right) \ln \left(1.75\right)$

$\setminus \rightarrow x \left(\ln \left(1.75\right) - \ln \left(3.5\right)\right) = 2 \ln \left(3.5\right) - 3 \ln \left(1.75\right)$

$\setminus \rightarrow x = \frac{\ln \left({3.5}^{2}\right) - \ln \left({1.75}^{3}\right)}{\ln \left(\frac{1.75}{3.5}\right)} = \ln \frac{\frac{16}{7}}{\ln} \left(\frac{1}{2}\right)$

$= \ln \frac{\frac{7}{16}}{\ln} \left(2\right) \approx - 1.19265$