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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