Uncovering the Mystery of the Lambert W-Function in Solving Equations

[MHD93]

Homework Statement

attached

The Attempt at a Solution

tried to divide the equation by 4^x and other attemptions, but they didn't work
supposing y = 1/2x didn't either work

 

[I like Serena]

I don't really see a way to solve this mathematically.

Numerically I get 2 roots:

1. -0.819769544935445
2. 0.729714973623184

but they don't look like anything recognizable.

 

[HallsofIvy]

$4^{x+1}- 4^{1/(2x)+ 1}- 2^x+ 1= 0$
Can be simplified by observing that $4^{x+1}= 4(4^x)= 4((2^2)^x)= 4((2^x)^2)$ and that $4^{1/(2x)+ 1}= 4(4^{1/2x})= 4((2^2)^{1/2x})= 4(2^{1/x})$

But it is that 1/x exponent that causes trouble.

 

[coolul007]

Mathematica says the equation reduces to this:

2^(2 + 1/x) == 1 +( 7) 2^x

It still has that pesky 1/x.

 

[MHD93]

Forgive me! please,, I was given the question wrong!
A little modification makes it so much easier that I solved it

$4^{x+1}- 4^{(1/2)x + 1}- 2^x + 1= 0$

This mistake was about to kill me!
I'm sorry
It's solved now, thanks!

 

[I like Serena]

Forgive me! please,, I was given the question wrong!
A little modification makes it so much easier that I solved it

$4^{x+1}- 4^{(1/2)x + 1}- 2^x + 1= 0$

This mistake was about to kill me!
I'm sorry
It's solved now, thanks!

Yes, that is a bit easier! :lol:

But the time was not wasted, I've learned a lot about the Lambert W-function, which I never heard of before this.