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

Feb 5, 2015

The answer is: $x = \frac{\ln 5 - \ln 2}{\ln 2 + 2 \ln 5}$.

The only way to find the solutions of this equation is to pass both members as arguments of the logarithmics function, better in base $e$.

So:

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

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

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

$x = \frac{\ln 5 - \ln 2}{\ln 2 + 2 \ln 5}$.