Solving a transcendental equation

1. Apr 15, 2012

geoduck

Suppose you have the equation

$$x^2*(log x)^{1/3}=-C$$

for very small C. I have a book claims that asymptotically, for very small C, the solution is:

$$x^2=C*\frac{2}{[logC]^{1/3} }$$

I'm not quite sure how to show this. If the 2 wasn't there, it looks like what they did was:

$$x^2=-\frac{C}{(log x)^{1/3}}$$

and just substituted -C for x on the RHS:

$$x^2=\frac{C}{(log C)^{1/3}}$$

I was thinking maybe you could substitute -C for x2 instead to get:

$$x^2=\frac{2^{1/3}C}{(log C)^{1/3}}$$

2. Apr 23, 2012

JJacquelin

I agree, there is a typo in the formula of the book : the power 1/3 is missing in the coefficient 2^(1/3)
An even better approximate can be derived (in attachment)

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