Problem 2
Question
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) \( \frac{1}{5} \cdot \log_{3}(y) \)."
1Step 1: Applying the Product Rule for Logarithms
The product rule for logarithms states that \( \log_b(AB) = \log_b(A) + \log_b(B) \). For the expression \( \log_{3}(xyz) \), apply the product rule iteratively: \[ \log_{3}(xyz) = \log_{3}(x) + \log_{3}(y) + \log_{3}(z) \]
2Step 2: Applying the Quotient Rule for Logarithms
The quotient rule for logarithms states that \( \log_b(\frac{A}{B}) = \log_b(A) - \log_b(B) \). For the expression \( \log_{3}(\frac{xz}{y}) \), apply this rule: \[ \log_{3}(\frac{xz}{y}) = \log_{3}(xz) - \log_{3}(y) \]Next, apply the product rule to \( \log_{3}(xz) \): \[ \log_{3}(xz) = \log_{3}(x) + \log_{3}(z) \]So, the expression becomes:\[ \log_{3}(\frac{xz}{y}) = \log_{3}(x) + \log_{3}(z) - \log_{3}(y) \]
3Step 3: Applying the Power Rule for Logarithms
The power rule for logarithms states that \( \log_b(A^n) = n \cdot \log_b(A) \). For the expression \( \log_{3}(\sqrt[5]{y}) \), note that \( \sqrt[5]{y} = y^{1/5} \). Apply the power rule:\[ \log_{3}(y^{1/5}) = \frac{1}{5} \cdot \log_{3}(y) \]
Key Concepts
Product Rule for LogarithmsQuotient Rule for LogarithmsPower Rule for Logarithms
Product Rule for Logarithms
The product rule for logarithms is a very handy tool when dealing with the logarithm of a product of numbers. It allows us to simplify this into a sum of individual logarithms. Essentially, this rule states that the logarithm of a product is equal to the sum of the logarithms of its factors. The formula is as follows:\[ \log_b(AB) = \log_b(A) + \log_b(B) \]This can be extended to more than two factors. For example, if you have three variables, say \(x, y,\) and \(z\), and you're trying to find \(\log_{3}(xyz)\), you would apply the product rule to find:\[ \log_{3}(xyz) = \log_{3}(x) + \log_{3}(y) + \log_{3}(z) \]This transformation makes larger expressions easier to handle as it breaks down the multiplicative components into simpler, additive parts.
- Useful for simplifying the multiplication of variables or expressions into separate logs.
- Makes calculations more straightforward when dealing with logarithms in algebra.
Quotient Rule for Logarithms
When you come across a division inside a logarithm, the quotient rule comes to the rescue. This rule states that the logarithm of a fraction is equivalent to the logarithm of the numerator minus the logarithm of the denominator, expressed in the following formula:\[ \log_b\left(\frac{A}{B}\right) = \log_b(A) - \log_b(B) \]Consider the expression \(\log_{3}\left(\frac{xz}{y}\right)\). By using the quotient rule, we first break it into:\[ \log_{3}(xz) - \log_{3}(y) \]Next, apply the product rule to further split \(\log_{3}(xz)\) into:\[ \log_{3}(x) + \log_{3}(z) \]Combining these steps, the complete expression becomes:\[ \log_{3}(x) + \log_{3}(z) - \log_{3}(y) \]
- Ideal for handling divisions within logarithms to simplify expressions before calculations.
- Transforms complex divisions into simpler problems by separating into subtraction of logs.
Power Rule for Logarithms
The power rule for logarithms simplifies expressions where a number is raised to an exponent within a logarithmic expression. This rule tells us that the exponent can be moved in front of the logarithm, multiplying it by the logarithm. It is expressed mathematically as:\[ \log_b(A^n) = n \cdot \log_b(A) \]Taking the expression \(\log_{3}(\sqrt[5]{y})\) as an example, recognize that a fifth root can be rewritten using exponents: \(y^{1/5}\). Hence, applying the power rule yields:\[ \log_{3}(y^{1/5}) = \frac{1}{5} \cdot \log_{3}(y) \]This allows for simplification by reducing the complexity of the original expression, making it more manageable for further mathematical operations.
- Enables easy manipulation of exponential terms within logarithms.
- Vital for solving equations involving powers within logarithmic contexts efficiently.
Other exercises in this chapter
Problem 1
Exer. 1-2: If possible, find (a) \(f^{-1}(5)\) and (b) \(g^{-1}(6)\) $$ \begin{aligned} &\begin{array}{|l|l|l|l|} \hline \boldsymbol{x} & 2 & 4 & 6 \\ \hline \b
View solution Problem 1
Solve the equation. $$7^{x+6}=7^{3 x-4}$$
View solution Problem 2
Exer. 1-2: Change to logarithmic form. (a) \(3^{5}=243\) (b) \(3^{-4}=\frac{1}{81}\) (c) \(c^{p}=d\) (d) \(7^{x}=100 p\) (e) \(3^{-2 x}=\frac{P}{F}\) (f) \((0.9
View solution Problem 2
Exer. 1-2: If possible, find (a) \(f^{-1}(5)\) and (b) \(g^{-1}(6)\) $$ \begin{aligned} &\begin{array}{|l|l|l|l|} \hline t & 0 & 3 & 5 \\ \hline f(t) & 2 & 5 &
View solution