Question 1042616
.
e^4x+5e^2x-24=0 

I have to solve and show work, I am completely lost
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


There are two ways.


<pre>
1.  Let y = {{{e^(2x)}}} be a new variable.

    Then your equation can be written in the form

    {{{y^2 + 5y - 24}}} = 0.

    Factor left side:

    (y + 8)*(y-3) = 0.

    The roots are y = -8 and y = 3.

    Then the original equation deploys in two:

    {{{e^(2x)}}} = -8,  which has no solutions,  and

    {{{e^(2x) -3}}} = 0, which is equivalent to {{{e^(2x)}}} = 3  and has the solution  2x = ln(3),  or  x = {{{ln(3)/2}}}.
</pre>

<pre>
2.  Factor the original equation as

    {{{(e^(2x)+8)*(e^(2x)-3)}}} = {{{0}}}.

    Then it deploys in two equations:

    {{{e^(2x)}}} = -8,  which has no solutions,  and

    {{{e^(2x)}}} = 3,  which . . . .    and so on from the n.1 . . . 
</pre>

Introducing new variable is a standard way and it makes the solution as simple as possible.


When and if you know this method well, you may apply direct factoring without introducing new variable.