How do you solve #e^(2x)-5e^x+6=0#?

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Answer 1 · 2017-01-13T15:41:54

#x = ln (2), ln (3)#

We have: #e^(2 x) - 5 e^(x) + 6 = 0#

Using the laws of exponents:

#Rightarrow (e^(x))^(2) - 5 e^(x) + 6 = 0#

Let's factorise the equation:

#Rightarrow (e^(x))^(2) - 2 e^(x) - 3 e^(x) + 6 = 0#

#Rightarrow e^(x) (e^(x) - 2) - 3 (e^(x) - 2) = 0#

#Rightarrow (e^(x) - 2) (e^(x) - 3) = 0#

#Rightarrow e^(x) = 2, 3#

Applying #ln# to both sides of the equation:

#Rightarrow ln (e^(x)) = ln (2), ln (3)#

Using the laws of logarithms:

#Rightarrow x ln (e) = ln (2), ln (3)#

#therefore x = ln (2), ln (3)#