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Write each expression as a single logarithm. $$\ln \sqrt{x-1}+\ln \sqrt{x+1}-2 \ln \left(x^{2}-1\right)$$

Short Answer

Expert verified
The expression as a single logarithm is \(-\frac{3}{2} \ln ((x-1)(x+1))\).

Step by step solution

01

Apply the Power Rule of Logarithms

The power rule of logarithms states that \( ext{n} \, ext{log} \, a = ext{log} \, a^n \). For example, \( 2 \ln A = \ln A^2 \). Let's apply this rule to the given expression, \(-2 \ln (x^2 - 1),\) turning it into \(\ln (x^2 - 1)^{-2}.\) The expression now becomes:\[\ln \sqrt{x-1} + \ln \sqrt{x+1} + \ln (x^2 - 1)^{-2}\]
02

Simplify Square Roots as Exponents

Recall that a square root can be expressed as an exponent of \( \frac{1}{2} \). Thus, \( \sqrt{x-1} = (x-1)^{1/2} \) and \( \sqrt{x+1} = (x+1)^{1/2} \). Substitute these into the expression:\[\ln (x-1)^{1/2} + \ln (x+1)^{1/2} + \ln (x^2 - 1)^{-2}\]
03

Apply the Product Rule of Logarithms

The product rule of logarithms states that \( \ln a + \ln b = \ln (a \times b) \). We have three logarithmic terms: \( \ln (x-1)^{1/2}, \ln (x+1)^{1/2}, \text{and} \ln (x^2 - 1)^{-2}\). These can be combined by multiplying their arguments:\[\ln \left[ (x-1)^{1/2} \cdot (x+1)^{1/2} \cdot (x^2-1)^{-2} \right]\]
04

Combine the Expressions Inside the Logarithm

Since \( (x-1)^{1/2} \cdot (x+1)^{1/2} = \sqrt{(x-1)(x+1)} \), and knowing that square roots can be expressed as \((x-1)(x+1)^{1/2},\) the expression becomes:\[\ln \left( \frac{\sqrt{(x-1)(x+1)}}{(x^2-1)^2} \right)\]Since \(x^2 - 1 = (x-1)(x+1)\), we can simplify further as \((x^2-1)^2 = ((x-1)(x+1))^2\), so:\[\ln \left( \frac{\sqrt{(x-1)(x+1)}}{((x-1)(x+1))^2} \right)\]
05

Simplification of Expression

We have \( \ln \left( \frac{\sqrt{(x - 1)(x + 1)}}{((x - 1)(x + 1))^2} \right) \), which simplifies to:\[\ln \left( (x - 1)^{1/2} \cdot (x + 1)^{1/2} \cdot \left((x - 1)\cdot(x + 1)\right)^{-2} \right)\]Simplifying inside the logarithm:\[\ln \left( \frac{1}{(x-1)^{3/2} \cdot (x+1)^{3/2}} \right)\]Since \( \frac{1}{A} = A^{-1} \), the expression becomes:\[\ln \left( ((x-1)(x+1))^{-3/2} \right)\]
06

Final Step: Write as a Single Logarithm

Putting it all together, the single logarithm expression is:\[-\frac{3}{2} \ln ((x-1)(x+1))\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule of Logarithms
Logarithms are incredibly handy when dealing with exponents, and the power rule of logarithms is a perfect illustration of this. It's a rule that states: for any positive number **a** and any real numbers **n**, \( n \cdot \log(a) = \log(a^n) \). This means exponentiation can be transformed into multiplication.

For example, if you have an expression such as \(2 \ln A\), it can be rewritten using the power rule as \(\ln A^2\). This is exactly what happens when we encounter the expression \(-2 \ln (x^2 - 1)\); it is rewritten using the power rule as \(\ln (x^2 - 1)^{-2}\).

This transformation is crucial because it simplifies dealing with expressions where variables are raised to powers, especially when working in the logarithmic domain.
Product Rule of Logarithms
Simplifying logarithmic expressions often involves using multiple logarithmic rules. One such rule is the product rule of logarithms. It states that the logarithm of a product is equal to the sum of the logarithms of the factors. In mathematical terms, this can be written as: \( \ln(a) + \ln(b) = \ln(a \times b) \).

This rule comes in handy when managing multiple logarithmic expressions like \(\ln (x-1)^{1/2} + \ln (x+1)^{1/2}\). By applying the product rule here, we can combine these into a single logarithmic term: \(\ln((x-1)^{1/2} \cdot (x+1)^{1/2})\).

This simplification approach is especially useful when aiming to express a combination of logarithms as a single logarithmic expression. Streamlining multiple logarithmic terms can make complex equations much easier to handle. So, understanding and applying the product rule of logarithms is a valuable skill in solving logarithmic problems.
Logarithmic Expressions
A logarithmic expression encompasses operations involving logarithms, such as addition, subtraction, and multiplication within the logarithmic scale. These expressions allow us to manipulate complex equations that involve exponents, roots, and products by transforming them into more manageable forms.

Considering the given exercise, where the aim is to write the expression \(\ln \sqrt{x-1}+\ln \sqrt{x+1}-2 \ln (x^{2}-1)\) as a single logarithm, it requires expert use of logarithmic laws.

We start by applying the power rule to handle the exponent, then convert square roots to exponential form, and use the product rule to combine elements under one log.As the last step, we rearrange and further simplify the expression, acknowledging that subtracting a logarithm can be represented as division inside the logarithm: turning terms like \(2 \ln (x^2 - 1)\) into dividing by \((x^2-1)^2\) under a single logarithm.

Understanding how to combine these logarithmic rules helps in reformulating and unraveling intricate expressions, making it easier to interpret and solve equations aesthetically and logically.

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Most popular questions from this chapter

Use the following formula for Newton’s Law of Cooling: If you take a hot dinner out of the oven and place it on the kitchen countertop, the dinner cools until it reaches the temperature of the kitchen. Likewise, a glass of ice set on a table in a room eventually melts into a glass of water at that room temperature. The rate at which the hot dinner cools or the ice in the glass melts at any given time is proportional to the difference between its temperature and the temperature of its surroundings (in this case, the room). This is called Newton's law of cooling (or warming) and is modeled by $$T=T_{S}+\left(T_{0}-T_{S}\right) e^{-k t}$$ where \(T\) is the temperature of an object at time \(t, T_{s}\) is the temperature of the surrounding medium, \(T_{0}\) is the temperature of the object at time \(t=0, t\) is the time, and \(k\) is a constant. A body is discovered in a hotel room. At 7: 00 A.M. a police detective found the body's temperature to be \(85^{\circ} \mathrm{F}\). At 8: 30 A.M. a medical examiner measures the body's temperature to be \(82^{\circ} \mathrm{F}\). Assuming the room in which the body was found had a constant temperature of \(74^{\circ} \mathrm{F}\), how long has the victim been dead? (Normal body temperature is \(98.6^{\circ} \mathrm{F} .\) ).

Evaluate the logarithms exactly (if possible). $$\log _{1 / 4} 4096$$

Write each logarithmic equation in its equivalent exponential form. $$y=\log _{x} z$$

State the domain, range, and \(x\) -intercept of the function \(f(x)=\log (a-x)-b\) for \(a\) and \(b\) real positive numbers.

Solve the exponential equations. Make sure to isolate the base to a power first. Round our answers to three decimal places. $$\left(3^{x}-3^{-x}\right)^{2}=0$$

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