- #1
geoduck
- 258
- 2
Suppose you have the equation
[tex]x^2*(log x)^{1/3}=-C [/tex]
for very small C. I have a book claims that asymptotically, for very small C, the solution is:
[tex]x^2=C*\frac{2}{[logC]^{1/3} } [/tex]
I'm not quite sure how to show this. If the 2 wasn't there, it looks like what they did was:
[tex]x^2=-\frac{C}{(log x)^{1/3}} [/tex]
and just substituted -C for x on the RHS:
[tex]x^2=\frac{C}{(log C)^{1/3}} [/tex]
I was thinking maybe you could substitute -C for x2 instead to get:[tex]x^2=\frac{2^{1/3}C}{(log C)^{1/3}} [/tex]
[tex]x^2*(log x)^{1/3}=-C [/tex]
for very small C. I have a book claims that asymptotically, for very small C, the solution is:
[tex]x^2=C*\frac{2}{[logC]^{1/3} } [/tex]
I'm not quite sure how to show this. If the 2 wasn't there, it looks like what they did was:
[tex]x^2=-\frac{C}{(log x)^{1/3}} [/tex]
and just substituted -C for x on the RHS:
[tex]x^2=\frac{C}{(log C)^{1/3}} [/tex]
I was thinking maybe you could substitute -C for x2 instead to get:[tex]x^2=\frac{2^{1/3}C}{(log C)^{1/3}} [/tex]