Question

Which expression is equivalent to log Subscript 8 Baseline 4 a (StartFraction b minus 4 Over c Superscript 4 Baseline EndFraction)?
log Subscript 8 Baseline 4 + log Subscript 8 Baseline a minus log Subscript 8 Baseline (b minus 4) minus 4 log Subscript 8 Baseline c
log Subscript 8 Baseline 4 + log Subscript 8 Baseline a + (log Subscript 8 Baseline (b minus 4) minus 4 log Subscript 8 Baseline c)
log Subscript 8 Baseline 4 a + log Subscript 8 Baseline b minus 4 minus 4 log Subscript 8 Baseline c minus 4
log Subscript 8 Baseline 4 + log Subscript 8 Baseline a minus log Subscript 8 Baseline (b minus 4) minus log Subscript 8 Baseline 4 c

Answer 1

Answer:

$\log _{8} 4+\log _{8} a+\log _{8}(b-4)-4 \log _{8} c$

The equivalent expression is log Subscript 8 Baseline 4 + log Subscript 8 Baseline a + (log Subscript 8 Baseline (b minus 4) minus 4 log Subscript 8 Baseline c)

Step-by-step explanation:

The expression is written as  $\log _{8}\left(4 a\left(\frac{b-4}{c^{4}}\right)\right)$

Using the log rule: $\log _{c}(x y)=\log _{c} x+\log _{c} y$, the expression can be written as,

$\log _{c}(x y)=\log _{c} x+\log _{c} y$

Using the log rule: $\log _{c}\left(\frac{x}{y}\right)=\log _{c} x-\log _{c} y$ , the expression can be written as,  

$\log _{8} 4+\log _{8} a+\log _{8}(b-4)-\log _{8} c^{4}$

Since, we know that, $\log _{a} x^{b}=b \log _{a} x$, the expression is written as,

$\log _{8} 4+\log _{8} a+\log _{8}(b-4)-4 \log _{8} c$

Thus, the equivalent expression is log Subscript 8 Baseline 4 + log Subscript 8 Baseline a + (log Subscript 8 Baseline (b minus 4) minus 4 log Subscript 8 Baseline c)

Answer 2

Answer:

Its b

Step-by-step explanation:

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