I'll give brainliest:
With step by step solution:
Solve for x in the equation;
X^2 = 16^x

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))

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