Sam is proving the product property of logarithms.
Which expression and justification completes the third step of her proof?
A. $\log _3(b^{x y})$; power rule of exponents
B. $\log _3(b^{x-y})$; subtraction property of exponents
C. $\log _3(3^{x+y})$; multiplication rule of exponents
D. $\log _a(b^{\frac{x}{y}})$; division property of exponents
Answer (1)
Apply the product of powers rule: b x \t ⋅ b y = b x + y .
Substitute b x + y into the logarithm: lo g 3 ( b x + y ) .
The third step is lo g 3 ( b x + y ) .
The correct expression and justification is: lo g 3 ( b x + y ) ; multiplication rule of exponents. lo g 3 ( b x + y ) ; m u lt i pl i c a t i o n r u l eo f e x p o n e n t s
Explanation
Analyzing the Given Information We are given the first two steps of Sam's proof of the product property of logarithms. We need to find the correct expression and justification for the third step. The proof starts with lo g 3 ( h M ) and substitutes h = b x and M = b y to get lo g 3 ( b x ⋅ b y ) .
Applying the Product of Powers Rule The next step involves simplifying the expression inside the logarithm. We know that when multiplying exponential terms with the same base, we add the exponents. That is, b x ⋅ b y = b x + y . Therefore, the next step in the proof is lo g 3 ( b x + y ) . The justification for this step is the product of powers rule.
Identifying the Correct Expression and Justification So, the third step of the proof is lo g 3 ( b x + y ) . Now we need to choose the correct option from the given choices. The correct expression is lo g 3 ( b x + y ) , and the justification is the multiplication rule of exponents, also known as the product of powers rule.
Final Answer Therefore, the completed step is:
\t lo g 3 ( b x + y ) ; multiplication rule of exponents
Examples
Logarithms are used in many real-world applications, such as calculating the magnitude of earthquakes on the Richter scale, measuring the loudness of sound in decibels, and determining the pH of a solution in chemistry. The product property of logarithms, which Sam is proving, allows us to simplify expressions involving logarithms of products, making these calculations easier. For example, if we want to find the logarithm of a large number that can be expressed as a product of smaller numbers, we can use the product property to break down the problem into simpler steps. This is particularly useful in fields like engineering and physics, where complex calculations are common.
