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Marcus is proving the Product Rule of Logarithms. The following is his work so far. Let logbm=x and logbn=y . logbm=x is equiva...Asked by 54t4n45
                Marcus is proving the Product Rule of Logarithms. The following is his work so far.
Let logbm=x
and logbn=y
.
logbm=x
is equivalent to bx=m
.
logbn=y
is equivalent to by=n
.
Therefore, mn=bxby
.
 
Which of the following options best describes the next step Marcus should take in his proof?
Option #1: Use the Product Rule of Exponents to rewrite the equation as mn=bx+y
.
Option #2: Rewrite this exponential equation as a logarithmic equation.
Option #3: Substitute specific values for m, n, x, and y.
(1 point)
The option that best describes the next step in the proof is Option #
            
            
        Let logbm=x
and logbn=y
.
logbm=x
is equivalent to bx=m
.
logbn=y
is equivalent to by=n
.
Therefore, mn=bxby
.
Which of the following options best describes the next step Marcus should take in his proof?
Option #1: Use the Product Rule of Exponents to rewrite the equation as mn=bx+y
.
Option #2: Rewrite this exponential equation as a logarithmic equation.
Option #3: Substitute specific values for m, n, x, and y.
(1 point)
The option that best describes the next step in the proof is Option #
Answers
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                    Answered by
            GPT-4o mini
            
    The best option for the next step in Marcus's proof is **Option #1**: Use the Product Rule of Exponents to rewrite the equation as \( mn = b^{x+y} \).
This follows logically because \( b^{x} \cdot b^{y} = b^{x+y} \) is the Product Rule of Exponents. Therefore, it’s the correct way to proceed from \( mn = b^{x} \cdot b^{y} \). The next logical step would be to relate this back to logarithms.
    
This follows logically because \( b^{x} \cdot b^{y} = b^{x+y} \) is the Product Rule of Exponents. Therefore, it’s the correct way to proceed from \( mn = b^{x} \cdot b^{y} \). The next logical step would be to relate this back to logarithms.
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