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lemon14 · Mar 19, 2011, 04:42 AM

$x^{lg x-1}=100$

galactus · Mar 19, 2011, 09:27 AM

Is this what you mean?:

$x^{log(x)-1}=100$

If so, look close. There are two solutions.One is rather obvious.

What is log(100)? Isn't it equal to 2?

So, $100^{log(100)-1}=100^{2-1}=\text{TA-DAA!}$

Can you find the other solution?

lemon14 · Mar 19, 2011, 10:18 AM

The solutions are 100 and 1/10. Thanks!

galactus · Mar 19, 2011, 10:25 AM

Yep. That's them. Good :)

galactus · Mar 19, 2011, 10:29 AM

But, if you want to do this algebraically, let''s do this:

Take log of both sides:

$log(x^{log(x)-1})=2$

$(log(x)-1)log(x)=2$

$(log(x))^{2}-log(x)-2=0$

Let u=log(x)

$u^{2}-u-2=0$

$u=-1, \;\ u=2$

Resub, since $u=log(x)$ :

$log(x)=-1, \;\ log(x)=2$

$x=\frac{1}{10}, \;\ x=100$