Solution
- The solution steps are given below:
[tex]\begin{gathered} \log_2(8)-\log_2(20)+\log_2(40) \\ \\ \text{ To solve the question, we need to apply the laws of logarithm} \\ \text{ The laws relevant for this question are:} \\ \log_ca+\log_cb=\log_c(a\times b)\text{ Law 1} \\ \log_ca-\log_cb=\log_c(\frac{a}{b})\text{ Law 2} \\ \log_aa^b=b\log_aa=b\text{ \lparen Since }\log_aa=1)\text{ Law 3} \\ \\ \\ \text{ Applying the laws, we have:} \\ \log_2(8)-\log_2(20)+\log_2(40) \\ \\ \text{ Applying Law 2,} \\ \log_2(\frac{8}{20})+\log_2(40) \\ \\ \text{ Applying Law 1,} \\ \log_2(\frac{8}{20}\times40) \\ \\ \text{ Simplify the expression} \\ \\ \log_2(8\times2) \\ \\ \log_2(16) \\ \\ \log_2(2^4) \\ \\ \text{ Applying Law 3,} \\ 4\log_22=4 \end{gathered}[/tex]
Final Answer
The answer is 4
