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Assoli18 [71]
2 years ago
7

[Quick Answer Needed] Which of the following shows the extraneous solution to the logarithmic equation?

Mathematics
1 answer:
Nitella [24]2 years ago
6 0

Answer:

C

Step-by-step explanation:

Given the logarithmic equation

\log_4x+\log_4(x-3)=\log_4(-7x+21)

First, notice that

x>0\\ \\x-3>0\Rightarrow x>3\\ \\-7x+21>0\Rightarrow 7x

So, there is no possible solutions, all possible solutions will be extraneous.

Solve the equation:

\log_4x+\log_4(x-3)=\log_4x(x-3),

then

\log_4x(x-3)=\log_4(-7x+21)\\ \\x(x-3)=-7x+21\\ \\x^2-3x+7x-21=0\\ \\x^2+4x-21=0\\ \\D=4^2-4\cdot 1\cdot (-21)=16+84=100\\ \\x_{1,2}=\dfrac{-4\pm 10}{2}=-7,\ 3

Hence, x=3 and x=-7 are extraneous solutions

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What is 8 3/5 + 4 4/5?
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Answer:

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Step-by-step explanation:

Step 1: <u>Define/explain.</u>

An easier way to solve this is by changing the mixed fractions to improper fractions.

To do this, multiply the whole number by the denominator, then add the product to the numerator; the denominator remains the same.

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From here, add as usual.

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Step 3: <u>Conclude.</u>

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To do this, divide the numerator by the denominator.

The amount of times the numerator evenly goes into the denominator is the whole number.

The amount of remaining numbers in the denominator.

The numerator remains the same.

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I, therefore, believe the answer to this is 13 2/5.

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