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