Log functions?

[ManicMechanic]

I'm assuming that you're working with an inequality, not an equation.

I got a numerical approximation by graphing y = ln x and y = x^1/3 superimposing those 2 graphs on the same grid. Where the ln x graph is above the x^1/3 graph, the x values for that region are the solution to the inequality ln x > x^1/3

If it really is an equation, look for the point of intersection of the 2 graphs.

A graphing calculator can give a pretty good approximation.

[maggiodd]

Not every equation can be solved algebraically. I suspect you can't solve this for x, but haven't played with it enough.

As previously suggested, you can get a numerical estimate as follows:
You have f(x)>g(x). Try graphing f(x)-g(x) and look for where the graph is above the x-axis. This portion of the graph is where f(x)>g(x). The x-intercept is where ln(x) = cuberoot(x). A graphing calculator can estimate that x-intercept.

Hope that helps.
Dan