Question

Which expression is equivalent to log Subscript w Baseline StartFraction (x squared minus 6) Superscript 4 Baseline Over RootIndex 3 StartRoot x squared + 8 EndRoot EndFraction?
4 log Subscript w Baseline StartFraction x squared Over 1296 EndFraction minus one-third log Subscript w Baseline (2 x + 8)
4 log Subscript w Baseline (x squared minus 6) minus one-third log Subscript w Baseline (x squared + 8)
4 log Subscript w Baseline (X squared minus 6) minus one-third log Subscript w Baseline (x squared + 8)
4 (log Subscript w Baseline x squared minus one-third log Subscript w Baseline (x squared + 8) minus 6)

Answer 1

Answer:

its c

Step-by-step explanation:

Answer 2

Option b: $4 \log _{w}\left(x^{2}-6\right)-\frac{1}{3} \log _{w}({x^{2}+8})$ is the correct answer.

Explanation:

The expression is $\log _{w}\left(\frac{\left(x^{2}-6\right)^{4}}{\sqrt[3]{x^{2}+8}}\right)$

Applying log rule, $\log _{c}\left(\frac{a}{b}\right)=\log _{c}(a)-\log _{c}(b)$, we get,

$\log _{w}\left(\left(x^{2}-6\right)^{4}\right)-\log _{w}(\sqrt[3]{x^{2}+8})$

Again applying the log rule, $\log _{a}\left(x^{b}\right)=b\cdot\log _{a}(x)$, we get,

$4 \log _{w}\left(x^{2}-6\right)-\log _{w}(\sqrt[3]{x^{2}+8})$

The cube root can be written as,

$4 \log _{w}\left(x^{2}-6\right)-\log _{w}({x^{2}+8})^{\frac{1}{3} }$

Applying the log rule, $\log _{a}\left(x^{b}\right)=b\cdot\log _{a}(x)$, we have,

$4 \log _{w}\left(x^{2}-6\right)-\frac{1}{3} \log _{w}({x^{2}+8})$

Thus, the expression which is equivalent to $\log _{w}\left(\frac{\left(x^{2}-6\right)^{4}}{\sqrt[3]{x^{2}+8}}\right)$ is $4 \log _{w}\left(x^{2}-6\right)-\frac{1}{3} \log _{w}({x^{2}+8})$

Hence, Option b is the correct answer.