Answer:
These types of questions again, where John K Williamson approximates again (No offense), and I answer with the question with Lambert W function again.
We’ll start with the following equation:
x2=16x=24x
Take the log of both sides:
2ln(x)=4xln(2)
Divide both sides by −2x
−ln(x)e−ln(x)=−2ln(2)
Use the Product Log (W-function):
−ln(x)=Wn(−2ln(2))
x=e−Wn(−2ln(2))
This is the first complex set of solutions! Apparently, none of them are real numbers, but weird, because we saw that negative half popping.
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We will do this following substitution (*):
x=−u or u=−x
Substitute into the equations:
u2=16−u=2−4u
We actually get an almost identical question, which leads to this:
−ln(u)e−ln(u)=2ln(2)
So we get:
u=e−Wn(2ln(2))
And substitute back x=−u
x=−e−Wn(2ln(2))
In particular, this is another set of solutions, which the negative half belongs to!
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Since:
2ln(2)=ln(2)eln(2)
Then:
W0(2ln(2))=ln(2)
And:
x=−e−ln(2)=−12
And we have it! A simple, real solution.
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Actually, all possible values of x are:
x=∓e−Wn(±2ln(2))
