User Portlet
| Discussions |
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| I guess based on definitions and propertied of square root, exponentials and logarithms ((x)^(2/3)) == ((x^2)^(1/3)) True for x>=0. ((Log[x])^2)^(1/3) == ((Log[x])^(1/3))^2 true for x>=1 & x |
User Portlet | Discussions |
|---|
| I guess based on definitions and propertied of square root, exponentials and logarithms ((x)^(2/3)) == ((x^2)^(1/3)) True for x>=0. ((Log[x])^2)^(1/3) == ((Log[x])^(1/3))^2 true for x>=1 & x |