# How do you solve  4^(x-1)=5^(x+1)?

May 31, 2015

${4}^{x} - 1 = {4}^{-} 1 \cdot {4}^{x} = {4}^{x} / 4$

${5}^{x + 1} = {5}^{1} \cdot {5}^{x} = 5 \cdot {5}^{x}$

Given

${4}^{x - 1} = {5}^{x + 1}$

We have

${4}^{x} / 4 = 5 \cdot {5}^{x}$

Multiply both sides by $4$ and divide both sides by ${5}^{x}$ to get

$20 = {4}^{x} / {5}^{x} = {\left(\frac{4}{5}\right)}^{x}$

Now take logs of both sides to get

$x \cdot \log \left(\frac{4}{5}\right) = \log \left(20\right)$

Divide both sides by $\log \left(\frac{4}{5}\right)$ to get

$x = \log \frac{20}{\log} \left(\frac{4}{5}\right) = \frac{\log \left(4\right) + \log \left(5\right)}{\log \left(4\right) - \log \left(5\right)}$

$\log \left(2\right) \cong 0.30103$

$\log \left(4\right) = 2 \log \left(2\right) \cong 0.60206$

$\log \left(5\right) = 1 - \log \left(2\right) \cong 0.69897$

$x \cong \frac{0.60206 + 0.69897}{0.60206 - 0.69897}$

$= \frac{1.30103}{-} 0.09691$

$\cong - 13.425$