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scrible · #11 November 7th, 2009, 07:54 AM

Quote:

Originally Posted by Defunkt

It's far simpler than you're making it out to be:

By the definition of a logarithm, $log_x(x)=1$ .

Why? -- we know that $x^{log_x(a)} = a$ . So $x^{log_x(x)} = x$ and therefore $log_x(x)=1$ .

Using this and the fact that $log_b(a^n) = n\cdot log_b(a)$ for any $a,b,n$ , we get:

$5\sqrt{x} =x + 12log_x(x^{\frac{1}{2}}) \Rightarrow 5\sqrt{x} = x + 12\cdot \frac{1}{2} \cdot \log_x(x) \Rightarrow$ $5\sqrt{x} = x + 6 \Rightarrow 25x = x^2 + 12x + 36 \Rightarrow x^2-13x+36=0$

I belive you can solve that.

Thanks a million. You know I keep going over the log laws, it is the application of the laws that is killing me. Do you know of any web sight with tricky examples like these?