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

May 22, 2015

One easy way is using logarithms because there is one law of logarithms which states that $\log {a}^{n} = n \cdot \log a$

So, let's put logarithms on both sides (to keep the equality):

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

Approximating $\log 5 \cong 0.7$ and $\log 4 \cong 0.6$, we get

$\left(x + 2\right) 0.7 = \left(1 - x\right) 0.6$
$0.7 x + 1.4 = 0.6 - 0.6 x$
$1.3 x = - 0.8$
$x = - \frac{0.8}{1.3} \cong - 0.61$

May 22, 2015

Another way to do that would be
${5}^{x + 2} = {4}^{1 - x}$
${5}^{x} \cdot {5}^{2} = {4}^{1} \cdot {4}^{- 1}$
$25 \cdot {5}^{x} = 4 \cdot \frac{1}{4} ^ x$
${5}^{x} \cdot {4}^{x} = \frac{4}{25}$
${20}^{x} = 0.16$
using the natural log
$\log {20}^{x} = \log 0.16$
$x \log 20 = \log 0.16$
$x = \frac{\log 0.16}{\log 20} \approx - 0.612$
or using the log base 20
${\log}_{20} {20}^{x} = {\log}_{20} 0.16$
$x = {\log}_{20} 0.16 \approx - 0.612$