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Log functions?
how do you solve for x when ln x> cube root of x
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Re: Log functions?
For any log function, you can use both sides of the equation as exponents for the base of the logarithm.
Spoiler for An Example:
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Re: Log functions?
I'm sorry, but I am still confused...what does that tell us? And your equation doesn't seem to be true...?
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Re: Log functions?
Quote:
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.) |
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. |
Re: Log functions?
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 |
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