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by **santaclaus** on Mon Apr 13, 2009 7:09 am

sq.rt (log base 3sq.rt.(x - log base 4 sq.rt (x))=1/2
so, I would start by simplifying what's inside the first square root
sq.rt (3 to some power =x and 4 to some power gives me x) and i'm stuck here b/c i can't really apply any log rules until I get x first? and then if I do know what x is, it's in the square root =1/2

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by **stapel_eliz** on Mon Apr 13, 2009 10:59 am

santaclaus wrote:sq.rt (log base 3sq.rt.(x - log base 4 sq.rt (x))=1/2

Please review this article on formatting, and reply with clarification and / or correction. I think you mean the equation to be as follows, but I'm not sure...?

. . . . . $\sqrt{\log_3\left(\sqrt{x\, -\, \log_4\left(\sqrt{x}\right)}\right)}\, =\, \frac{1}{2}$

When you reply, please include a clear listing of your steps and reasoning. Thank you!

The Purplemath Forums

solve log equations with sq.rt

solve log equations with sq.rt