```Question 209731
{{{e^(2x) - 3e^x + 2 = 0}}}
{{{e^(2x) - 3e^x = -2}}}
Not wrong, but doesn't help.
e^((2x)/(x^3)) = e^(-2)
???! There is no way to arrive at this statement from the previous one.<br>
The key to solving the equation, {{{e^(2x) - 3e^x + 2 = 0}}}, is to recognize that<ul><li>{{{e^(2x) = (e^x)^2}}}</li><li>the equation can then be written as: {{{(e^x)^2 - 3e^x + 2 = 0}}}</li><li>the equation is therefore a quadratic equation in {{{e^x}}}</li></ul>
If you have trouble seeing the last two, we can use a substitution to make it clearer. Let {{{q = e^x}}}. Replacing {{{e^x}}} with q we get:
{{{q^2 - 3q + 2 = 0}}}.<br>
We can solve this for q (aka {{{e^x}}}) with the quadratic formula or by factoring:
{{{(q - 2)(q - 1) = 0}}}
In order for this product to be zero, one of the factors must be zero:
{{{q - 2 = 0}}} or {{{q - 1 = 0}}}
Solving these we get:
{{{q = 2}}} or {{{q = 1}}}
Of course we are not interested in "q". We are interested in "x". So we can substitute back in for "q":
{{{e^x = 2}}} or {{{e^x = 1}}}
To solve these for x we will find the natural log of each side:
{{{ln(e^x) = ln(2)}}} or {{{ln(e^x) = ln(1)}}}
Using one of the properties of logarithms: {{{log(a, (x^y)) = y*log(a, x)}}} we get:
x*ln(e) = ln(2) or x*ln(e) = ln(1)
Since the ln(e) is 1 by definition and ln(1) is zero:
x = ln(2) or x = 0```