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Simplifying Exponential Equations Using Lambert W FunctionThe Basic IdeaThe W function is defined as the inverse of f(x)=x ex, which means
You can think of it as a kind of "log" function. In fact, Mathematica calls the function "ProductLog". Starting with the definition z = W(z) eW(z), you can divide both sides by W(z), giving you this useful identity:
This is a handy factoid, because sometimes (as we will see) we get a solution that looks like e^W(something), which now we know is the same as something/W(something). We'll use this in the last step of example 2, below. To solve equations involving exponents, the trick is to get the variables all on the right side, and then manipulate the right side until the equation looks something like
Then you "take the W of both sides", giving you
Example 1: Solve for t, 2t=5tTo solve the equation 2t=5t, we divide by 2t to get
then multiply by -ln(2)/5 to get
Now the right hand side is in that special form that looks like b eb, so we "take the W of both sides" giving us
Example 2: Solve for x, xx=16For this problem, which I will generalize to xx=z, the trick is to express x as eln(x). Here are the steps:
So the solution to xx=16 is ln(16)/W(ln(16)), or about 2.74536802356746. Example 3: Solve iz=z
This is a problem that comes up in connection with the value of i^i^i^... As is our custom, let's solve the more general problem, a^z=z, where a is any constant.
So, for our problem, in which -ln(i) = -i pi/2,
Approximate value: z = 0.4382829367270321 + 0.3605924718713855i Internet References
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