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

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Answer 1 · 2015-12-13T12:42:59

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

We will use the following properties:

#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#