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Chapter 4: Problem 2

Exer. 1-8: Express in terms of logarithms of \(x, y, z,\) or \(w\) (a) \(\log _{3}(x y z)\) (b) \(\log _{3}(x z / y)\) (c) \(\log _{3} \sqrt[5]{y}\)

Short Answer

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(a) \(\log_3 x + \log_3 y + \log_3 z\), (b) \(\log_3 x + \log_3 z - \log_3 y\), (c) \(\frac{1}{5} \log_3 y\)."

Step by step solution

01

Applying the Product Rule for Logarithms for Part (a)

To express \(\log_3(xyz)\) in terms of \(\log_3 x\), \(\log_3 y\), and \(\log_3 z\), use the product rule for logarithms: \(\log_b(uvw) = \log_b u + \log_b v + \log_b w\). Here, \(u = x\), \(v = y\), and \(w = z\). Therefore, \(\log_3(xyz) = \log_3 x + \log_3 y + \log_3 z\).
02

Using the Quotient Rule for Logarithms for Part (b)

To express \(\log_3(xz/y)\), use the quotient rule for logarithms: \(\log_b(u/v) = \log_b u - \log_b v\). Here, \(u = xz\) and \(v = y\). Thus, \(\log_3(xz/y) = \log_3(xz) - \log_3 y\). Now apply the product rule to \(\log_3(xz) = \log_3 x + \log_3 z\). Therefore, \(\log_3(xz/y) = (\log_3 x + \log_3 z) - \log_3 y\).
03

Applying the Power Rule for Logarithms for Part (c)

To express \(\log_3 \sqrt[5]{y}\), use the power rule for logarithms: \(\log_b u^n = n\log_b u\). Here, \(u = y\), and \(n = \frac{1}{5}\). Thus, \(\log_3 \sqrt[5]{y} = \log_3(y^{1/5}) = \frac{1}{5} \log_3 y\).

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Key Concepts

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

Product Rule for Logarithms
The product rule for logarithms is an essential tool in simplifying expressions involving the products of variables. The rule states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. Mathematically, it can be expressed as:
  • \(\log_b(uvw) = \log_b u + \log_b v + \log_b w\)
This rule is particularly useful because it transforms a more complex logarithmic expression into a simple sum.
For example, consider the expression \(\log_3(xyz)\). Using the product rule, we break down this expression into a sum of three separate logarithms: \(\log_3 x + \log_3 y + \log_3 z\).
This approach makes computations and algebraic manipulations much more manageable.
Quotient Rule for Logarithms
The quotient rule for logarithms helps simplify expressions involving the division of two quantities. According to this rule, the logarithm of a quotient is equal to the difference between the logarithms of the numerator and the denominator:
  • \(\log_b(u/v) = \log_b u - \log_b v\)
This rule allows us to easily deal with logarithmic expressions that involve division.
Take the expression \(\log_3(xz/y)\) as an example. Using the quotient rule, we can write this as \(\log_3(xz) - \log_3 y\). By further applying the product rule to \(\log_3(xz)\), we find that \(\log_3(xz) = \log_3 x + \log_3 z\). Thus, the whole expression becomes \(\log_3 x + \log_3 z - \log_3 y\).
This step-by-step simplification helps in better understanding and solving logarithmic identities with ease.
Power Rule for Logarithms
The power rule for logarithms provides a powerful method to deal with expressions involving powers. This rule states that the logarithm of a number raised to an exponent is equal to the product of the exponent and the logarithm of the base number:
  • \(\log_b u^n = n\log_b u\)
This is particularly helpful when simplifying expressions that include roots or powers.
For instance, to simplify \(\log_3 \sqrt[5]{y}\), recognize that \(\sqrt[5]{y}\) is equivalent to \(y^{1/5}\). Using the power rule, this becomes \(\frac{1}{5} \log_3 y\). The power rule significantly reduces the complexity of logarithmic expressions by allowing us to move exponents to the front, turning a power into a factor.
Having mastered these rules ensures you can handle a variety of logarithmic problems with confidence.

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Most popular questions from this chapter

According to Newton's law of cooling, the rate at which an object cools is directly proportional to the difference in temperature between the object and the surrounding medium. The face of a household iron cools from \(125^{\circ}\) to \(100^{\circ}\) in 30 minutes in a room that remains at a constant temperature of \(75^{\circ} .\) From calculus, the temperature \(f(t)\) of the face after \(t\) hours of cooling is given by \(f(t)=50(2)^{-2 t}+75\) (a) Assuming \(t=0\) corresponds to 1: 00 PM., approximate to the nearest tenth of a degree the temperature of the face at 2: 00 P.M., 3: 30 P.M., and 4: 00 P.M. (b) Sketch the graph of \(f\) for \(0 \leq t \leq 4\)

Approximate the function at the value of \(x\) to four decimal places. (a) \(f(x)=\log \left(2 x^{2}+1\right)-10^{-x}, \quad x=1.95\) (b) \(g(x)=\frac{x-3.4}{\ln x+4}, \quad x=0.55\)

For manufacturers of computer chips, it is important to consider the fraction \(F\) of chips that will fail after \(t\) years of service. This fraction can sometimes be approximated by the formula \(F=1-e^{-c t},\) where \(c\) is a positive constant. (a) How does the value of \(c\) affect the reliability of a chip? (b) If \(c=0.125,\) after how many years will 35 \(\%\) of the chips have failed?

Graph \(f,\) and estimate its zeros. $$f(x)=x^{3} e^{x}-x^{2} e^{2 x}+1$$

Cooling A jar of boiling water at \(212^{\circ} \mathrm{F}\) is set on a table in a room with a temperature of \(72^{\circ} \mathrm{F}\). If \(T(t)\) represents the temperature of the water after \(t\) hours, graph \(T(t)\) and determine which function best models the situation. (1) \(T(t)=212-50 t\) (2) \(T(t)=140 e^{-t}+72\) (3) \(T(t)=212 e^{-t}\) (4) \(T(t)=72+10 \ln (140 t+1)\)
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