Question

which equation is equivalent to \\(\log_{3}(2x^{4} + 8x^{3}) - 3\log_{3}x = 2\log_{3}x\\)?\
\\(\circ\\) \\(\log_{3}(-x^{3} + 8x^{2}) = \log_{3}x^{2}\\)\
\\(\circ\\) \\(-2\log_{3}(2x^{4} + 8x^{3} - x) = \log_{3}x^{2}\\)\
\\(\circ\\) \\(\log_{3}(2x + 8) = \log_{3}x^{2}\\)

Explanation:

Step1: Apply log power rule

Recall the logarithm power rule: \( n\log_b a=\log_b a^n \). So, \( 3\log_3 x = \log_3 x^3 \) and \( 2\log_3 x=\log_3 x^2 \). The original equation is \( \log_3(2x^4 + 8x^3)-\log_3 x^3=\log_3 x^2 \).

Step2: Apply log quotient rule

Recall the logarithm quotient rule: \( \log_b a-\log_b c = \log_b\frac{a}{c} \). So, \( \log_3\frac{2x^4 + 8x^3}{x^3}=\log_3 x^2 \).

Step3: Simplify the fraction

Simplify \( \frac{2x^4 + 8x^3}{x^3} \). Factor out \( 2x^3 \) from the numerator: \( \frac{2x^3(x + 4)}{x^3} \). Cancel out \( x^3 \) (assuming \( x
eq0 \)), we get \( 2x + 8 \). So the equation becomes \( \log_3(2x + 8)=\log_3 x^2 \).

Answer:

\( \log_3(2x + 8) = \log_3x^2 \) (the third option)