x^{lg x-1}=100

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?

The solutions are 100 and 1/10. Thanks!

Yep. That's them. Good :)

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