Asked by Anonymous
                Whenever w is an integer greater than 1, log(w) w^2 / w^6 =?
F. -4
G. -3
H. -1 / 3
J. 1 / 3
K. 3
            
        F. -4
G. -3
H. -1 / 3
J. 1 / 3
K. 3
Answers
                    Answered by
            Reiny
            
    looking at the choices of answer I must conclude that you meant:
log<sub>w</sub> (w^2/w^6)
= log<sub>w</sub> w^2 - log<sub>w</sub> (w^6)
= 2 log<sub>w</sub>w/(6log<sub>w</sub>w)
= 2/6
= 1/3
    
log<sub>w</sub> (w^2/w^6)
= log<sub>w</sub> w^2 - log<sub>w</sub> (w^6)
= 2 log<sub>w</sub>w/(6log<sub>w</sub>w)
= 2/6
= 1/3
                    Answered by
            Anonymous
            
    wouldn't the power be -4? 2 - 6
    
                    Answered by
            Anthony
            
    Hello people from 2015. Take me back. Please
    
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