Problem 2
Question
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) \(\log _{3} \sqrt[5]{y}\)
Step-by-Step Solution
Verified Answer
(a) \(\log_3 x + \log_3 y + \log_3 z\), (b) \(\log_3 x + \log_3 z - \log_3 y\), (c) \(\frac{1}{5} \log_3 y\)."
1Step 1: 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, \(\log_3(xyz) = \log_3 x + \log_3 y + \log_3 z\).
2Step 2: 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, \(\log_3(xz/y) = (\log_3 x + \log_3 z) - \log_3 y\).
3Step 3: 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, \(\log_3 \sqrt[5]{y} = \log_3(y^{1/5}) = \frac{1}{5} \log_3 y\).
Key Concepts
Product Rule for LogarithmsQuotient Rule for LogarithmsPower Rule for Logarithms
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:
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.
- \(\log_b(uvw) = \log_b u + \log_b v + \log_b w\)
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:
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.
- \(\log_b(u/v) = \log_b u - \log_b v\)
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:
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.
- \(\log_b u^n = n\log_b u\)
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.
Other exercises in this chapter
Problem 1
Find \(\begin{array}{ll}\text { (a) } f^{-1}(5) & \text { (b) } g^{-1}(6)\end{array}\) $$\begin{array}{|c|c|c|c|} \hline x & 2 & 4 & 6 \\ \hline f(x) & 3 & 5 &
View solution Problem 2
Change to logarithmic form. (a) \(3^{3}=243\) (b) \(3^{-4}=\frac{1}{81}\) (c) \(c^{r}=d\) (d) \(7^{x}=100 p\) (e) \(3^{-2 x}=\frac{P}{F}\) (f) \((0.9)^{y}=\frac
View solution Problem 2
Solve the equation. $$6^{7-x}=6^{2 x+1}$$
View solution Problem 2
Find \(\begin{array}{ll}\text { (a) } f^{-1}(5) & \text { (b) } g^{-1}(6)\end{array}\) $$\begin{array}{|l|l|ll|} \hline t & 0 & 3 & 5 \\ \hline f(t) & 2 & 5 & 6
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