text: # How do you solve e^(-x) = 5^(2x) ?
  
  Dec 13, 2015
  
  Apply properties of logarithms to find that $x = 0$
  
  #### Explanation:
  
  We will use the following properties:
  
  • $\ln \left({a}^{x}\right) = x \ln \left(a\right)$
  
  • $\ln \left(e\right) = 1$
  
  ${e}^{- x} = {5}^{2 x}$
  
  $\implies \ln \left({e}^{- x}\right) = \ln \left({5}^{2 x}\right)$
  
  $\implies - x \ln \left(e\right) = 2 x \ln \left(5\right)$
  
  $\implies - x = 2 \ln \left(5\right) x$
  
  $\implies 2 \ln \left(5\right) x + x = 0$
  
  $\implies x \left(2 \ln \left(5\right) + 1\right) = 0$
  
  $\implies x = 0$