Turnbull Thornton
08/21/2023 · Elementary School
\( \leftarrow \) 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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Step-by-step Solution
To combine the logarithmic expressions into a single logarithm, we can use the properties of logarithms. Specifically, we will use the following properties:
1. \( \ln(a) + \ln(b) = \ln(ab) \)
2. \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)
Let's start with the given expression:
\[
\ln \left(\frac{x}{x-1}\right) + \ln \left(\frac{x+1}{x}\right) - \ln \left(x^{2}-1\right)
\]
First, we can combine the first two logarithms using the first property:
\[
\ln \left(\frac{x}{x-1}\right) + \ln \left(\frac{x+1}{x}\right) = \ln \left(\frac{x}{x-1} \cdot \frac{x+1}{x}\right)
\]
This simplifies to:
\[
\ln \left(\frac{x(x+1)}{x(x-1)}\right) = \ln \left(\frac{x+1}{x-1}\right)
\]
Now, we can substitute this back into the original expression:
\[
\ln \left(\frac{x+1}{x-1}\right) - \ln \left(x^{2}-1\right)
\]
Next, we apply the second property to combine these two logarithms:
\[
\ln \left(\frac{\frac{x+1}{x-1}}{x^{2}-1}\right)
\]
Now, we need to simplify the expression inside the logarithm. Recall that \( x^2 - 1 \) can be factored as \( (x-1)(x+1) \):
\[
\ln \left(\frac{x+1}{x-1} \cdot \frac{1}{(x-1)(x+1)}\right) = \ln \left(\frac{x+1}{(x-1)(x+1)}\right)
\]
This simplifies to:
\[
\ln \left(\frac{1}{x-1}\right)
\]
Thus, the final expression as a single logarithm is:
\[
\ln \left(\frac{1}{x-1}\right) = -\ln(x-1)
\]
So, the final answer is:
\[
\ln \left(\frac{1}{x-1}\right) = -\ln(x-1)
\]
Quick Answer
\[
\ln \left(\frac{1}{x-1}\right) = -\ln(x-1)
\]
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