# How do you solve 5^(x+2) = 4^(x+1)?

Jan 9, 2016

$x = \frac{\ln 4 - 2 \ln 5}{\ln 5 - \ln 4}$

#### Explanation:

Here is another way to solve this exponential equation:

${5}^{x + 2} = {4}^{x + 1}$

Take the natural log of both sides...

$\ln {5}^{x + 2} = \ln {4}^{x + 1}$

Use log rules, distribution, and combine like terms...

$\left(x + 2\right) \ln 5 = \left(x + 1\right) \ln 4$

$x \ln 5 + 2 \ln 5 = x \ln 4 + \ln 4$

$x \left(\ln 5 - \ln 4\right) = \ln 4 - 2 \ln 5$

$x = \frac{\ln 4 - 2 \ln 5}{\ln 5 - \ln 4}$

Jan 9, 2016

$x = \frac{\ln 4 - 2 \ln 5}{\ln 5 - \ln 4}$

#### Explanation:

Take the logarithm with base $5$ of both sides.

$x + 2 = {\log}_{5} \left({4}^{x + 1}\right)$

Rewrite using logarithm rules.

$x + 2 = \left(x + 1\right) {\log}_{5} 4$

Rewrite ${\log}_{5} 4$ with the change of base formula.

$x + 2 = \frac{\left(x + 1\right) \ln 4}{\ln} 5$

$\left(x + 2\right) \ln 5 = \left(x + 1\right) \ln 4$

$x \ln 5 + 2 \ln 5 = x \ln 4 + \ln 4$

$x \ln 5 - x \ln 4 = \ln 4 - 2 \ln 5$

$x \left(\ln 5 - \ln 4\right) = \ln 4 - 2 \ln 5$

$x = \frac{\ln 4 - 2 \ln 5}{\ln 5 - \ln 4}$

As has been displayed, logarithmic equations are exceptionally versatile and can be manipulated in many ways to receive the same result.

Note that this process could be repeated by taking the logarithm with base $4$ of both sides instead of $5$.

Feb 7, 2016

$x = \ln \left({\left(\frac{2}{5}\right)}^{\frac{2}{5}}\right)$

#### Explanation:

${5}^{x + 2} = {4}^{x + 1}$

$25 \cdot {5}^{x} = 4 \cdot {4}^{x}$

${5}^{x} = \left(\frac{4}{25}\right) \cdot {4}^{x}$

$\ln \left({5}^{x}\right) = \ln \left(\left(\frac{4}{25}\right) \cdot {4}^{x}\right)$

$x \ln 5 = \ln \left(\frac{4}{25}\right) + x \ln 4$

$x \left(\ln 5\right) - x \ln \left(4\right) = \ln \left(\frac{4}{25}\right)$

$x \ln \left(\frac{5}{4}\right) = \ln \left(\frac{4}{25}\right)$

$x = \ln \left(\frac{4}{25}\right) : \ln \left(\frac{5}{4}\right)$

$x = \ln \left({\left(\frac{4}{25}\right)}^{\frac{4}{5}}\right)$
$x = \ln \left({\left(\frac{2}{5}\right)}^{\frac{2}{5}}\right)$