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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