Answer:
[tex]3 \log 2[/tex]
Step-by-step explanation:
Given expression:
[tex]\log 2 + \log \left(\dfrac{3}{2}\right)+\log \left(\dfrac{4}{3}\right)+\log \left(\dfrac{5}{4}\right)+\log \left(\dfrac{6}{5}\right)+\log \left(\dfrac{7}{6}\right)+\log \left(\dfrac{8}{7}\right)[/tex]
[tex]\textsf{Apply the \underline{Log Product Law}}: \quad \log_ax + \log_ay=\log_axy[/tex]
[tex]\implies \log \left(2 \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot \dfrac{5}{4} \cdot \dfrac{6}{5} \cdot \dfrac{7}{6} \cdot \dfrac{8}{7}\right)[/tex]
Cross out common factors:
[tex]\implies \log \left(\diagup\!\!\!\!2 \cdot \dfrac{\diagup\!\!\!\!3}{\diagup\!\!\!\!2} \cdot \dfrac{\diagup\!\!\!\!4}{\diagup\!\!\!\!3} \cdot \dfrac{\diagup\!\!\!\!5}{\diagup\!\!\!\!4} \cdot \dfrac{\diagup\!\!\!\!6}{\diagup\!\!\!\!5} \cdot \dfrac{\diagup\!\!\!\!7}{\diagup\!\!\!\!6} \cdot \dfrac{8}{\diagup\!\!\!\!7}\right)[/tex]
Therefore:
[tex]\implies \log 8[/tex]
Factor the number: 8 = 2³
[tex]\implies \log 2^3[/tex]
[tex]\textsf{Apply the \underline{Log Power Law}}: \quad \log_ax^n=n\log_ax[/tex]
[tex]\implies 3 \log 2[/tex]
