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Let \(f(x)=\ln \left(1-\frac{1}{x^{2}}\right)\) (a) Use the properties of logarithms (and some algebra) to show that$$f(2)+f(3)+f(4)=\ln \frac{5}{8}$$ (b) Use a calculator to check the result in part (a).

Short Answer

Expert verified
The sum of \( f(2) + f(3) + f(4) = \ln \frac{5}{8} \).

Step by step solution

01

Apply Properties of Logarithms

For the function \( f(x) = \ln \left(1 - \frac{1}{x^2} \right) \), we know the property of logarithms: \( \ln a + \ln b + \ln c = \ln(abc) \). Therefore, \( f(2) + f(3) + f(4) = \ln \left( \left(1 - \frac{1}{4} \right) \left(1 - \frac{1}{9} \right) \left(1 - \frac{1}{16} \right) \right) \).
02

Simplify the Expressions

Simplify each term: \( 1 - \frac{1}{4} = \frac{3}{4} \), \( 1 - \frac{1}{9} = \frac{8}{9} \), and \( 1 - \frac{1}{16} = \frac{15}{16} \). Multiply these expressions: \( \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} \).
03

Calculate the Product

Calculate the product of these fractions: \( \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} = \frac{3 \times 8 \times 15}{4 \times 9 \times 16} = \frac{360}{576} \). Simplify this fraction to \( \frac{5}{8} \).
04

Apply Logarithm to the Simplified Fraction

Now apply the logarithm: \( \ln \left( \frac{5}{8} \right) \). Therefore, \( f(2) + f(3) + f(4) = \ln \frac{5}{8} \).
05

Verify Using Calculator

Check each value using a calculator: Calculate \( f(2) = \ln(\frac{3}{4}) \), \( f(3) = \ln(\frac{8}{9}) \), \( f(4) = \ln(\frac{15}{16}) \). Then verify that their sum equals \( \ln(\frac{5}{8}) \), confirming the manual calculation.

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Key Concepts

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

Logarithmic Functions
Logarithmic functions are powerful mathematical tools used to solve various types of equations, especially those involving exponential growth and decay. In this exercise, the function under consideration is given by \( f(x) = \ln \left(1 - \frac{1}{x^2} \right) \). Here, \( \ln \) represents the natural logarithm, which is logarithm with the base \( e \), a mathematical constant approximately equal to 2.718.
To simplify expressions like this, it's crucial to utilize the properties of logarithms. One fundamental property is that \( \ln a + \ln b = \ln(ab) \). In simple terms, this means that the logarithm of a product can be split into the sum of the logarithms. In the given exercise, this property allows us to combine the separate logarithmic terms, \( f(2), f(3), \) and \( f(4) \) into a single expression of \( \ln \left(\frac{5}{8}\right) \).
By understanding and applying these properties, solving logarithmic equations becomes more straightforward and less intimidating.
Fraction Multiplication
Fraction multiplication is an essential arithmetic operation used frequently when simplifying various mathematical expressions, including those involving logarithms. Here, we need to multiply several fractions as part of the simplification process in the problem.
Working through our example, consider the fractions involved: \( \frac{3}{4} \), \( \frac{8}{9} \), and \( \frac{15}{16} \). Multiplying fractions involves multiplying the numerators together and the denominators together:
  • For the numerators: \( 3 \times 8 \times 15 = 360 \).
  • For the denominators: \( 4 \times 9 \times 16 = 576 \).
As a result, we have the fraction \( \frac{360}{576} \). It's important to simplify this resulting fraction to its lowest terms by dividing both the numerator and the denominator by their greatest common divisor, which is 72 in this case, resulting in \( \frac{5}{8} \). Perfectly executing fraction multiplication and simplification ensures a correct and simplified logarithmic result.
Simplification of Expressions
Simplifying mathematical expressions involves reducing them into their simplest forms, making them easier to work with and understand. In the context of the given problem, simplification is achieved through a series of steps utilizing the properties of logarithms and fraction multiplication.
The initial expressions \( 1 - \frac{1}{4} \), \( 1 - \frac{1}{9} \), and \( 1 - \frac{1}{16} \) can be directly evaluated to produce simpler fractions: \( \frac{3}{4} \), \( \frac{8}{9} \), and \( \frac{15}{16} \), respectively. Each expression in this stage is simplified by executing basic subtraction and finding a common fraction form.
Subsequently, when we have to handle the fraction multiplication that follows: \( \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} \), the role of simplification is emphasized again. The process culminates as these values are worked through, showing that the multiplication of simplified fractions aligns with logarithmic combination rules, leading to \( \ln \left(\frac{5}{8} \right) \). Mastery of simplification techniques is vital for solving complex problems efficiently, as it reduces the complexity and potential for error while calculating.

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Most popular questions from this chapter

The following extract is from an article by Kim Murphy that appeared in the Los Angeles Times on September 14 1994 CAIRO-Over a chorus of reservations from Latin America and Islamic countries still troubled about abortion and family issues, nearly 180 nations adopted a wide-ranging plan Tuesday on global population, the first in history to obtain partial endorsement from the Vatican. The plan, approved on the final day of the U.N. population conference here, for the first time tries to limit the growth of the world's population by preventing it from exceeding 7.2 billion people over the next two decades. (a) In 1995 the world population was 5.7 billion, with a relative growth rate of \(1.6 \% /\) year. Assuming continued exponential growth at this rate, make a projection for the world population in the year \(2020 .\) Round off the answer to one decimal place. How does your answer compare to the target value of 7.2 billion mentioned in the article? (b) As in part (a), assume that in 1995 the world population was 5.7 billion. Determine a value for the growth constant \(k\) so that exponential growth throughout the years \(1995-2020\) leads to a world population of 7.2 billion in the year 2020 .

(a) Let \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t} .\) Show that \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)=e^{k}-1 .\) (This is actually done in detail in the text. So, ideally, you should look back only if you get stuck or want to check your answer.) (b) Assume as given the following approximation, which was introduced in Exercise 26 of Section 5.2 \(e^{x} \approx x+1 \quad\) provided \(x\) is close to zero Use this approximation to explain why \(e^{k}-1 \approx k\) provided that \(k\) is close to zero. Remark: Combining this result with that in part (a), we conclude that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is approximately equal to the growth constant \(k .\) As explained in the text, this is one of the reasons why in applications we've not distinguished between the relative growth rate and the decay constant \(k\)

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