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:

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

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

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