Thread: Log functions?
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Unread 21-10-2008, 20:34
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Re: Log functions?

Quote:
Originally Posted by Nickzdabomb View Post
And your equation doesn't seem to be true...?
It's not an equation, and neither is yours. They're both inequalities.

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.)
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