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Math  /  Algebra

QuestionChoose the expression that shows an expanded form of the logarithm. ln⁡(x10x−13x−20)(12)ln⁡x10(x−1)−ln⁡(3x−20)10ln⁡x+(12)ln⁡(x−1)−ln⁡(3x−20)10ln⁡x+ln⁡x−1−ln⁡(3x−12)\begin{array}{c} \ln \left(\frac{x^{10} \sqrt{x-1}}{3 x-20}\right) \\ \left(\frac{1}{2}\right) \ln x^{10}(x-1)-\ln (3 x-20) \\ 10 \ln x+\left(\frac{1}{2}\right) \ln (x-1)-\ln (3 x-20) \\ 10 \ln x+\ln \sqrt{x-1}-\ln (3 x-12) \end{array}

Studdy Solution

STEP 1

1. We are given a logarithmic expression and asked to find its expanded form.
2. We will use logarithmic properties to expand the expression.
3. The properties of logarithms include: - ln⁡(a⋅b)=ln⁡a+ln⁡b\ln(a \cdot b) = \ln a + \ln b - ln⁡(ab)=ln⁡a−ln⁡b\ln\left(\frac{a}{b}\right) = \ln a - \ln b - ln⁡(ab)=bln⁡a\ln(a^b) = b \ln a

STEP 2

1. Apply the quotient rule for logarithms.
2. Apply the product rule for logarithms.
3. Apply the power rule for logarithms.

STEP 3

Apply the quotient rule for logarithms to the expression:
ln⁡(x10x−13x−20)=ln⁡(x10x−1)−ln⁡(3x−20)\ln \left(\frac{x^{10} \sqrt{x-1}}{3x-20}\right) = \ln(x^{10} \sqrt{x-1}) - \ln(3x-20)

STEP 4

Apply the product rule for logarithms to the numerator:
ln⁡(x10x−1)=ln⁡(x10)+ln⁡(x−1)\ln(x^{10} \sqrt{x-1}) = \ln(x^{10}) + \ln(\sqrt{x-1})

STEP 5

Apply the power rule for logarithms to each term:
ln⁡(x10)=10ln⁡x\ln(x^{10}) = 10 \ln x
ln⁡(x−1)=ln⁡((x−1)1/2)=12ln⁡(x−1)\ln(\sqrt{x-1}) = \ln((x-1)^{1/2}) = \frac{1}{2} \ln(x-1)

STEP 6

Combine the results from the previous steps:
ln⁡(x10x−13x−20)=10ln⁡x+12ln⁡(x−1)−ln⁡(3x−20)\ln \left(\frac{x^{10} \sqrt{x-1}}{3x-20}\right) = 10 \ln x + \frac{1}{2} \ln(x-1) - \ln(3x-20)
The expanded form of the logarithm is:
10ln⁡x+12ln⁡(x−1)−ln⁡(3x−20)10 \ln x + \frac{1}{2} \ln(x-1) - \ln(3x-20)

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