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Re: Log functions?
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.
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