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

3 answers

looking at the choices of answer I must conclude that you meant:

logw (w^2/w^6)
= logw w^2 - logw (w^6)
= 2 logww/(6logww)
= 2/6
= 1/3
wouldn't the power be -4? 2 - 6
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