Sanders Dunn
05/16/2023 · Junior High School
Write the expression as a single logarithm. Express powers as factors. \( \ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right) \) \( \ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right)=\square \)
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To write the given expression as a single logarithm, we can use the properties of logarithms. The properties we will use are:
1. \( \ln(a) + \ln(b) = \ln(ab) \)
2. \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)
Given expression: \( \ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right) \)
Using the properties mentioned above, we can simplify the expression as follows:
\( \ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right) \)
\( = \ln \left(\frac{x}{x-1} \cdot \frac{x+1}{x}\right) - \ln \left(x^{2}-1\right) \)
\( = \ln \left(\frac{x(x+1)}{x(x-1)}\right) - \ln \left(x^{2}-1\right) \)
\( = \ln \left(\frac{x+1}{x-1}\right) - \ln \left(x^{2}-1\right) \)
Therefore, the expression \( \ln \left(\frac{x}{x-1}\right)+\ln \left(\frac{x+1}{x}\right)-\ln \left(x^{2}-1\right) \) can be written as a single logarithm as \( \ln \left(\frac{x+1}{x-1}\right) - \ln \left(x^{2}-1\right) \).
Quick Answer
\( \ln \left(\frac{x+1}{x-1}\right) - \ln \left(x^{2}-1\right) \)
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