Question 1

Select all the correct answers.Which expressions are equivalent to the given expression?510810 I + 10810 20 - 10810 10

Question 2

Answer:

Options 1 and 4.

Explanation:

Given the expression:

$5\log_(10)x+\log_(10)20-\log_(10)10$

First, we can rewrite 20 as a product of 2 and 10.

$\begin{gathered} 5\operatorname{\log}_(10)x+\operatorname{\log}_(10)20-\operatorname{\log}_(10)10=5\operatorname{\log}x+\operatorname{\log}_(10)(2*10)-\operatorname{\log}_(10)10 \\ \text{ By the product law of logarithm: }log(a* b)=loga+logb \\ =5\operatorname{\log}_(10)x+\operatorname{\log}_(10)2+\operatorname{\log}_(10)10-\operatorname{\log}_(10)10 \\ =5\operatorname{\log}_(10)x+\operatorname{\log}_(10)2 \\ \text{ By the power law of logarithms: }nlogx=\log x^n \\ =\operatorname{\log}_(10)x^5+\operatorname{\log}_(10)2 \\ \text{ Applying the product law:} \\ =\operatorname{\log}_(10)(2x^5) \end{gathered}$

This is equivalent to Option 1.

$\begin{gathered} 5\operatorname{\log}_(10)x+\operatorname{\log}_(10)20-\operatorname{\log}_(10)10 \\ \text{Apply}\imaginaryI\text{ng the power law: }n\log x=\log x^n \\ =\operatorname{\log}x^5+\operatorname{\log}_(10)20-\operatorname{\log}_(10)10 \\ \text{ By the product law of logarithm: }log(a* b)=loga+logb \\ =\operatorname{\log}_(10)(20x^5)-\operatorname{\log}_(10)10 \\ By\text{ the unity law of logarithms: }\log_aa=1 \\ =\operatorname{\log}_(10)(20x^5)-1 \end{gathered}$

This is equivalent to Option 4.

Options 1 and 4 are the equivalent options.

0 Comments

Your comment on this question:

Your answer

1 Answer

0 Comments

Your comment on this answer:

Other Questions