Question

Hey help me on this??

Answer 1

The answer is C

Explanation:

Use such properties of logarithms:

\log_w\dfrac{a}{b}=\log_wa-\log_wb,\\ \\\log_wa^b=b\log_wa

Thus,

\log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2+8} }=\text{use the first property}=\\ \\=\log_w(x^2-6)^4-\log_w\sqrt[3]{x^2+8}=\\ \\=\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}=\text{use the second property}=\\ \\=4\log(x^2-6)-\dfrac{1}{3}\log_w(x^2+8).

Answer 2

C

Explanation:

Use such properties of logarithms:

`\log_w(a)/(b)=\log_wa-\log_wb,\\ \\\log_wa^b=b\log_wa`

Thus,

`\log_w\frac{(x^2-6)^4}{\sqrt[3]{x^2+8} }=\text{use the first property}=\\ \\=\log_w(x^2-6)^4-\log_w\sqrt[3]{x^2+8}=\\ \\=\log_w(x^2-6)^4-\log_w(x^2+8)^{(1)/(3)}=\text{use the second property}=\\ \\=4\log(x^2-6)-(1)/(3)\log_w(x^2+8).`