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
wouldn't the power be -4? 2 - 6
Hello people from 2015. Take me back. Please
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
To simplify the given expression, we can use the properties of logarithms and exponents. Let's break it down step by step:
1. Start with the expression: log(w) (w^2 / w^6).
2. Apply the quotient rule of logarithms: log(w) (w^2) - log(w) (w^6).
3. Simplify the exponents: log(w) (w^2) - log(w) (1 / w^4).
4. Use the power rule of logarithms: 2 log(w) (w) - 4 log(w) (1).
5. Apply the logarithm identity: log(w) (w) = 1.
6. Substitute the values: 2(1) - 4(0).
7. Simplify: 2 - 0 = 2.
Therefore, the value of log(w) (w^2 / w^6) is 2.
Since none of the answer choices provided match our result, it appears that the given expression has been simplified incorrectly or may contain a mistake. Please double-check the expression or verify if there are any additional instructions or context provided.