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

1 Answer
Dec 13, 2015

Apply properties of logarithms to find that #x=0#

Explanation:

We will use the following properties:

  • #ln(a^x) = xln(a)#

  • #ln(e) = 1#


#e^(-x) = 5^(2x)#

#=> ln(e^(-x)) = ln(5^(2x))#

#=> -xln(e) = 2xln(5)#

#=> -x = 2ln(5)x#

#=> 2ln(5)x+x = 0#

#=> x(2ln(5)+1) = 0#

#=> x = 0#