Answer:
Options 1 and 4.
Explanation:
Given the expression:
[tex]5\log_{10}x+\log_{10}20-\log_{10}10[/tex]First, we can rewrite 20 as a product of 2 and 10.
[tex]\begin{gathered} 5\operatorname{\log}_{10}x+\operatorname{\log}_{10}20-\operatorname{\log}_{10}10=5\operatorname{\log}x+\operatorname{\log}_{10}(2\times10)-\operatorname{\log}_{10}10 \\ \text{ By the product law of logarithm: }log(a\times b)=loga+logb \\ =5\operatorname{\log}_{10}x+\operatorname{\log}_{10}2+\operatorname{\log}_{10}10-\operatorname{\log}_{10}10 \\ =5\operatorname{\log}_{10}x+\operatorname{\log}_{10}2 \\ \text{ By the power law of logarithms: }nlogx=\log x^n \\ =\operatorname{\log}_{10}x^5+\operatorname{\log}_{10}2 \\ \text{ Applying the product law:} \\ =\operatorname{\log}_{10}(2x^5) \end{gathered}[/tex]This is equivalent to Option 1.
Next:
[tex]\begin{gathered} 5\operatorname{\log}_{10}x+\operatorname{\log}_{10}20-\operatorname{\log}_{10}10 \\ \text{Apply}\imaginaryI\text{ng the power law: }n\log x=\log x^n \\ =\operatorname{\log}x^5+\operatorname{\log}_{10}20-\operatorname{\log}_{10}10 \\ \text{ By the product law of logarithm: }log(a\times b)=loga+logb \\ =\operatorname{\log}_{10}(20x^5)-\operatorname{\log}_{10}10 \\ By\text{ the unity law of logarithms: }\log_aa=1 \\ =\operatorname{\log}_{10}(20x^5)-1 \end{gathered}[/tex]This is equivalent to Option 4.
Options 1 and 4 are the equivalent options.
