Here is another comment regarding Zeph's questions. I really enjoy answering questions. It sorts of review me about everything i learned last year.
Zeph,
Now I'm going to show you how to do question 16. Special credits to benofschool for showing me how to do this.
[log_3(X)]^2 - log_3(x)^2 = 3
Apply the rules of logarithms.
[log_3(x)]^2 - 2log_3(x) = 3
Let, log_3(x)= y
y^2-2y=3
Apply algebraic massage
y^2-2y-3=0
Solve for y.
(y+1) (y-3)=0
y=-1 y=3
Substitute back log_3(x)= y
log_3(x)=-1 and log_3(x)=3
x= 1/3 and 27
I'm going to post the answer for the other questions tomorrow.
-m@rk
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