SOLUTION: e^4x+5e^2x-24=0 I have to solve and show work, I am completely lost

Algebra.Com
Question 1042616: e^4x+5e^2x-24=0
I have to solve and show work, I am completely lost
Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.
e^4x+5e^2x-24=0
I have to solve and show work, I am completely lost
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are two ways. 

1.  Let y =  be a new variable.

    Then your equation can be written in the form

     = 0.

    Factor left side:

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

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

    Then the original equation deploys in two:

     = -8,  which has no solutions,  and

     = 0, which is equivalent to  = 3  and has the solution  2x = ln(3),  or  x = .


2.  Factor the original equation as

     = .

    Then it deploys in two equations:

     = -8,  which has no solutions,  and

     = 3,  which . . . .    and so on from the n.1 . . .

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.


Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
e^4x+5e^2x-24=0
Treat it as a quadratic with e^2x=x
x^2+5x-24=0
(x+8)(x-3)=0
(e^2x+8)(e^2x-3)=0
e^2x=-8; ln of both sides means ln -8, and that doesn't exist, so ignore that root.
e^2x=3, setting the factor equal to 0 and adding 3 to both sides
Take the ln of both sides. ln e^2x=2x, since ln and e cancel each other out.
2x=ln 3
x=(1/2)ln 3, exact answer. The numerical answer is 0.5493

-----------
e^2ln3+5e^ln3-24;
but e^2ln3=e^ln(9) by property of logs
e^ln9+5 e^ln3-24=9+15-24=0.
RELATED QUESTIONS

I am supposed to solve the following linear system using linear combinations. Please help (answered by Fombitz)
I am completely lost on how to work this problem. solve: y-2-8sqrt(y-2)+15 = 0 (answered by scott8148)
Solve: (x^4)-(8x^3)+(19x^2)-(16x)+(34)=0... Given that 4-i is a root Show your work... (answered by Fombitz)
So I have to solve for the direct variation and then graph -2y = 5x, and I am completely... (answered by ikleyn,Theo)
i need help to solve this. I am completely... (answered by Poohbear272)
solve the system, show steps please 2x+y+3=0 x^2+y^2=5 i keep geting x=plus or... (answered by MathLover1)
I am trying to solve the equation for x^2 – 2x – 13 = 0 I have to use the Indian Method... (answered by KMST)
I am trying to solve this problem & I am completely lost. Divide:... (answered by richwmiller)
Can I please get help solving these equations to find the value of x and can I please see (answered by greenestamps)