Log functions?

[Nickzdabomb]

how do you solve for x when ln x> cube root of x

[EricH]

For any log function, you can use both sides of the equation as exponents for the base of the logarithm.


In the case ln(x)< x^2, it would look something like e^ln(x) > e^x^2, solve as far as you can go or as far as the teacher wants.

[Nickzdabomb]

I'm sorry, but I am still confused...what does that tell us? And your equation doesn't seem to be true...?

[EricH]

And your equation doesn't seem to be true...?
It's not an equation, and neither is yours. They're both inequalities.:D

Now, e^ln (anything) = anything. If you haven't been taught that yet, then you should have been. So, e^ln (x) = x. You get x > whatever you've set ln(x) greater than.

Your original equation (if it can be called that) reads, "ln x> cube root of x", or, ln(x) > x^1/3. Now, e^ln(x)>e^x^1/3.

If that was an equation, then it can be solved. (It does seem to be a difficult one, or maybe I'm just a bit rusty.)

[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