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Chapter 5: Problem 26

Write the quantity using sums and differences of simpler logarithmic expressions. Express the answer so that logarithms of products, quotients, and powers do not appear. (a) \(\log _{b} \sqrt[3]{\frac{(x-1)^{2}(x-2)}{(x+2)^{2}(x+1)}}\) (b) \(\ln \left(\frac{e-1}{e+1}\right)^{3 / 2}\)

Short Answer

Expert verified
(a) \( \frac{1}{3} (2\log_{b}(x-1) + \log_{b}(x-2) - 2\log_{b}(x+2) - \log_{b}(x+1)) \) (b) \( \frac{3}{2} \ln(e-1) - \frac{3}{2} \ln(e+1) \)

Step by step solution

01

Simplify the expression inside the logarithm (Part a)

Rewrite the expression inside the logarithm by expanding it based on the rules of exponents. The cube root can be rewritten using exponents:\[ \sqrt[3]{\frac{(x-1)^{2}(x-2)}{(x+2)^{2}(x+1)}} = \left(\frac{(x-1)^{2}(x-2)}{(x+2)^{2}(x+1)}\right)^{1/3} \]
02

Apply the power rule (Part a)

Use the power rule for logarithms, which states that \( \log_{b}(a^{c}) = c \cdot \log_{b}(a) \):\[ \log_{b} \left(\frac{(x-1)^{2}(x-2)}{(x+2)^{2}(x+1)}\right)^{1/3} = \frac{1}{3} \log_{b}\left(\frac{(x-1)^{2}(x-2)}{(x+2)^{2}(x+1)}\right) \]
03

Apply the quotient rule (Part a)

Use the quotient rule for logarithms, which states that \( \log_{b} \left( \frac{m}{n} \right) = \log_{b}(m) - \log_{b}(n) \):\[ \frac{1}{3} \left( \log_{b}((x-1)^{2}(x-2)) - \log_{b}((x+2)^{2}(x+1)) \right) \]
04

Apply the product rule (Part a)

Use the product rule for logarithms, \( \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \), to simplify each part:\[ \log_{b}((x-1)^{2}(x-2)) = \log_{b}((x-1)^{2}) + \log_{b}(x-2) \]\[ \log_{b}((x+2)^{2}(x+1)) = \log_{b}((x+2)^{2}) + \log_{b}(x+1) \]
05

Apply the power rule to individual terms (Part a)

Simplify each logarithm containing powers using the power rule:\[ \log_{b}((x-1)^{2}) = 2\log_{b}(x-1) \]\[ \log_{b}((x+2)^{2}) = 2\log_{b}(x+2) \]Now substitute these back into the expression:\[ \frac{1}{3} \left( 2\log_{b}(x-1) + \log_{b}(x-2) - 2\log_{b}(x+2) - \log_{b}(x+1) \right) \]
06

Simplify the expression inside the logarithm (Part b)

Rewrite the expression inside the logarithm using the power rule from the beginning:\[ \ln \left( \frac{e-1}{e+1} \right)^{3/2} = \frac{3}{2} \ln \left( \frac{e-1}{e+1} \right) \]
07

Apply the quotient rule (Part b)

Use the quotient rule for logarithms to simplify the expression:\[ \frac{3}{2} \left( \ln(e-1) - \ln(e+1) \right) \]
08

Final Expression (Part a)

The simplified expression for part (a) is:\[ \frac{1}{3} \left( 2\log_{b}(x-1) + \log_{b}(x-2) - 2\log_{b}(x+2) - \log_{b}(x+1) \right) \]
09

Final Expression (Part b)

The simplified expression for part (b) is:\[ \frac{3}{2} \ln(e-1) - \frac{3}{2} \ln(e+1) \]

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Key Concepts

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

Power Rule for Logarithms
The power rule for logarithms is a handy tool for simplifying expressions where a term inside the logarithm is raised to a power. The rule states:
  • For logarithms of any base, the power rule is expressed as \( \log_{b}(a^{c}) = c \cdot \log_{b}(a) \).
When you encounter a logarithm of a power, the exponent can be pulled out to the front as a multiplier. This is especially useful for making expressions simpler and more manageable.
For example, consider the expression \( \log_{b}(x^3) \). According to the power rule, this can be rewritten as \( 3 \cdot \log_{b}(x) \).
Using this rule, you can transform complex logarithmic expressions into sums or differences of simpler ones, making calculations or further simplifications easier to handle.
Quotient Rule for Logarithms
The quotient rule for logarithms is crucial when you're dealing with a logarithm of a division between two terms. This rule can transform a logarithm of a quotient into a difference between two separate logarithms. It is represented as:
  • \( \log_{b} \left( \frac{m}{n} \right) = \log_{b}(m) - \log_{b}(n) \).
With the quotient rule, complicated expressions can be broken down. For instance, if you have \( \log_{b} \left( \frac{x}{y} \right) \), it becomes \( \log_{b}(x) - \log_{b}(y) \).
This transformation into a difference is beneficial for computations, especially when solving equations or further simplifying logarithmic expressions. You can focus on the individual components rather than tackling them all at once in a single fraction.
Product Rule for Logarithms
The product rule for logarithms is a powerful method for simplifying expressions involving the logarithm of a product. Whenever you see a logarithm of a multiplication of two terms, this rule helps break it down into a sum of separate logarithms. The product rule can be expressed as:
  • \( \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \).
This is especially helpful when dealing with complicated expressions, as it allows you to split the terms into simpler parts. For example, if you have \( \log_{b}(xy) \), using the product rule, you can rewrite it as \( \log_{b}(x) + \log_{b}(y) \).
By breaking down the expression, calculations become more straightforward, and it's easier to identify and manage each component separately. This rule is commonly used in algebra and calculus for making complex manipulations possible.

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Most popular questions from this chapter

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$4^{5-x}>15$$

(a) Graph the two functions \(f(x)=(\ln x) /(\ln 3)\) and \(g(x)=\ln x-\ln 3 .\) (Use a viewing rectangle in which \(x\) extends from 0 to 10 and \(y\) extends from \(-5 \text { to } 5 .)\) Why aren't the two graphs identical? That is, doesn't one of the basic log identities say that \((\ln a) /(\ln b)=\ln a-\ln b ?\) (b) Your picture in part (a) indicates that $$\frac{\ln x}{\ln 3}>\ln x-\ln 3 \quad(0< x \leq 10)$$ Find a viewing rectangle in which \((\ln x) /(\ln 3) \leq \ln x-\ln 3.\) (c) Use the picture that you obtain in part (b) to estimate the value of \(x\) for which \((\ln x) /(\ln 3)=\ln x-\ln 3.\) (d) Solve the equation \((\ln x) /(\ln 3)=\ln x-\ln 3\) algebraically and use the result to check your estimate in part (c).

Solve the equation \(\log _{2} x=\log _{x} 3 .\) For each root, give an exact expression and a calculator approximation rounded to two decimal places.

Radiocarbon Dating: Because rubidium-87 decays so slowly, the technique of rubidium-strontium dating is generally considered effective only for objects older than 10 million years. In contrast, archeologists and geologists rely on the radiocarbon dating method in assigning ages ranging from 500 to 50,000 years. Two types of carbon occur naturally in our environment: carbon-12, which is nonradioactive, and carbon-14, which has a half-life of 5730 years. All living plant and animal tissue contains both types of carbon, always in the same ratio. (The ratio is one part carbon- 14 to \(10^{12}\) parts carbon-12.) As long as the plant or animal is living, this ratio is maintained. When the organism dies, however, no new carbon-14 is absorbed, and the amount of carbon-14 begins to decrease exponentially. since the amount of carbon-14 decreases exponentially, it follows that the level of radioactivity also must decrease exponentially. The formula describing this situation is $$\mathcal{N}=\mathcal{N}_{0} e^{k T}$$ where \(T\) is the age of the sample, \(\mathcal{N}\) is the present level of radioactivity (in units of disintegrations per hour per gram of carbon), and \(\mathcal{N}_{0}\) is the level of radioactivity \(T\) years ago, when the organism was alive. Given that the half-life of carbon-14 is 5730 years and that \(\mathcal{N}_{0}=920\) disintegrations per hour per gram, show that the age \(T\) of a sample is given by $$T=\frac{5730 \ln (\mathcal{N} / 920)}{\ln (1 / 2)}$$

Prove that \(\left(\log _{a} x\right) /\left(\log _{a b} x\right)=1+\log _{a} b\)
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