In Exercises 109–112, find the domain of each logarithmic function. f(x) = log[(x+1)/(x-5)]
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Introduction to Logarithms
Problem 67
Textbook Question
The figure shows the graph of f(x) = ln x. In Exercises 65–74, use transformations of this graph to graph each function. Graph and give equations of the asymptotes. Use the graphs to determine each function's domain and range.

h(x) = ln (2x)
Verified step by step guidance1
Start with the base function \(f(x) = \ln x\), which has a vertical asymptote at \(x = 0\), domain \((0, \infty)\), and range \((-\infty, \infty)\).
The function \(h(x) = \ln(2x)\) represents a horizontal scaling of the base function by a factor of \(\frac{1}{2}\) inside the argument of the logarithm.
To find the new vertical asymptote, set the inside of the logarithm equal to zero: \$2x = 0\(, which gives \)x = 0\(. So the vertical asymptote remains at \)x = 0$.
Determine the domain of \(h(x)\) by solving \$2x > 0\(, which simplifies to \)x > 0\(. Thus, the domain is \)(0, \infty)$, same as the original function.
The range of \(h(x)\) remains \((-\infty, \infty)\) because logarithmic functions are continuous and unbounded vertically regardless of horizontal scaling.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Logarithmic Functions and Their Graphs
A logarithmic function, such as f(x) = ln(x), is the inverse of an exponential function. Its graph passes through (1,0) and has a vertical asymptote at x = 0. The function is defined only for positive x-values, giving it a domain of (0, ∞) and a range of all real numbers.
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Transformations of Functions
Transformations involve shifting, stretching, compressing, or reflecting a graph. For h(x) = ln(2x), the factor 2 inside the logarithm compresses the graph horizontally by a factor of 1/2. This changes the domain and shifts the vertical asymptote accordingly, affecting the graph's shape and position.
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Asymptotes and Domain of Logarithmic Functions
The vertical asymptote of a logarithmic function occurs where the argument of the log equals zero. For h(x) = ln(2x), the asymptote is at x = 0 since 2x = 0 when x = 0. The domain is all x-values making the argument positive, so here the domain is (0, ∞). Understanding asymptotes helps define where the function is valid.
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Introduction to Asymptotes
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