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Magazine Chapter 5: Problem 93
Given that \(\log _{a} x=2, \log _{a} y=3,\) and \(\log _{a} z=4\) find $$\log _{a} \frac{\sqrt[4]{y^{2} z^{5}}}{\sqrt[4]{x^{3} z^{-2}}}$$
Short Answer
Expert verified
The value is 7.
Step by step solution
01
Understand the logarithmic properties
To solve the expression, recall the properties of logarithms. These include product, quotient, and power rules. Specifically: 1. \(\text{log}_a (x y) = \text{log}_a x + \text{log}_a y\)2. \(\text{log}_a \frac{x}{y} = \text{log}_a x - \text{log}_a y\)3. \(\text{log}_a (x^k) = k \text{log}_a x\)
02
Simplify the given expression
Rewrite the given expression \(\log_a \frac{\sqrt[4]{y^2 z^5}}{\sqrt[4]{x^3 z^{-2}}}\) using the properties of radicals. Recall that the fourth root of a number is the same as raising that number to the power of \(\frac{1}{4}\). So, \(\frac{\sqrt[4]{y^2 z^5}}{\sqrt[4]{x^3 z^{-2}}} = \frac{(y^2 z^5)^{\frac{1}{4}}}{(x^3 z^{-2})^{\frac{1}{4}}}\).
03
Apply the exponent rule
Apply the exponent rule: \(\frac{(y^2 z^5)^{\frac{1}{4}}}{(x^3 z^{-2})^{\frac{1}{4}}} = \frac{y^{\frac{1}{2}} z^{\frac{5}{4}}}{x^{\frac{3}{4}} z^{-\frac{1}{2}}}\). Note that \(z^{-\frac{1}{2}}\) brings the z term in the denominator to the numerator as \(z^{\frac{1}{2}}\).
04
Combine like terms
Combine the z terms in the numerator using the property \(x^a x^b = x^{a + b}\). This gives \(\frac{y^{\frac{1}{2}} z^{\frac{5}{4}+\frac{1}{2}}}{x^{\frac{3}{4}}} = \frac{y^{\frac{1}{2}} z^{\frac{7}{4}}}{x^{\frac{3}{4}}}\).
05
Use the quotient property of logarithms
Rewrite the logarithm of a quotient as the difference of logarithms: \(\log_a \frac{y^{\frac{1}{2}} z^{\frac{7}{4}}}{x^{\frac{3}{4}}} = \log_a (y^{\frac{1}{2}} z^{\frac{7}{4}}) - \log_a (x^{\frac{3}{4}})\).
06
Apply the power rule of logarithms
Use the power rule for each term: \(\log_a (y^{\frac{1}{2}} z^{\frac{7}{4}}) - \log_a (x^{\frac{3}{4}}) = \frac{1}{2} \log_a y + \frac{7}{4} \log_a z - \frac{3}{4} \log_a x\).
07
Substitute the given logarithmic values
Given \(\log_a x = 2\), \(\log_a y = 3\), and \(\log_a z = 4\), substitute these values into the expression: \(\frac{1}{2}(3) + \frac{7}{4}(4) - \frac{3}{4}(2) = \frac{3}{2} + 7 - \frac{3}{2}\).
08
Simplify the expression
Combine like terms to get the final result: \(\frac{3}{2} + 7 - \frac{3}{2} = 7\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
logarithms
Logarithms are a way to express exponents. In the statement \( \text{log}_a x \), \( a \) is the base, and \( x \) is the result of raising \( a \) to some power. For example, \( \text{log}_2 8 = 3 \) because \( 2^3 = 8 \). Logarithms help solve equations where the variable is an exponent. This is crucial in many fields such as engineering, computer science, and mathematics.
They provide a way to transform multiplicative problems into additive ones, simplifying complex calculations.
The notation \( \text{log} \) may be used with different bases, such as common logarithms (base 10) often written as \( \text{log}_{10} \) or natural logarithms (base \( e \) ) written as \( \text{ln} \).
Understanding logarithms lays a foundation for exploring exponential and logarithmic functions, which are prevalent in more advanced mathematical studies.
They provide a way to transform multiplicative problems into additive ones, simplifying complex calculations.
The notation \( \text{log} \) may be used with different bases, such as common logarithms (base 10) often written as \( \text{log}_{10} \) or natural logarithms (base \( e \) ) written as \( \text{ln} \).
Understanding logarithms lays a foundation for exploring exponential and logarithmic functions, which are prevalent in more advanced mathematical studies.
logarithmic rules
Logarithmic rules tell us how to manipulate logarithms in various expressions. Three essential logarithm properties are:
1. **Product Rule:** \( \text{log}_a (xy) = \text{log}_a x + \text{log}_a y \).
This rule shows that the log of a product is the sum of the logs.
2. **Quotient Rule:** \( \text{log}_a \frac{x}{y} = \text{log}_a x - \text{log}_a y \).
This demonstrates that the log of a quotient is the difference between the logs.
3. **Power Rule:** \( \text{log}_a (x^k) = k \text{log}_a x \).
This indicates that the log of a power is the exponent multiplied by the log.
These rules are foundational and are used to simplify and solve logarithmic equations. In our exercise, each step relies on these basic properties to transform and simplify the expression. For instance, changing radicals into exponents or breaking up a fraction into the difference of logs. This theoretical background is key to understanding how to work with logarithmic expressions in practical problems.
1. **Product Rule:** \( \text{log}_a (xy) = \text{log}_a x + \text{log}_a y \).
This rule shows that the log of a product is the sum of the logs.
2. **Quotient Rule:** \( \text{log}_a \frac{x}{y} = \text{log}_a x - \text{log}_a y \).
This demonstrates that the log of a quotient is the difference between the logs.
3. **Power Rule:** \( \text{log}_a (x^k) = k \text{log}_a x \).
This indicates that the log of a power is the exponent multiplied by the log.
These rules are foundational and are used to simplify and solve logarithmic equations. In our exercise, each step relies on these basic properties to transform and simplify the expression. For instance, changing radicals into exponents or breaking up a fraction into the difference of logs. This theoretical background is key to understanding how to work with logarithmic expressions in practical problems.
precalculus
Precalculus is the study that prepares students for calculus. It covers various crucial topics including functions, complex numbers, and trigonometry.
A significant part of precalculus is understanding logarithms and their properties, which are essential for solving exponential equations and transforming complex expressions.
By mastering logarithms in precalculus, students build a foundation for calculus, where they will encounter more advanced topics like the natural logarithm and its applications in differentiation and integration.
The logarithmic function is a pivotal concept that shows up in growth models, decay processes, and other real-world contexts. This is why deep comprehension and application in problems, like the one discussed earlier, is critical before advancing further in mathematical studies.
A significant part of precalculus is understanding logarithms and their properties, which are essential for solving exponential equations and transforming complex expressions.
By mastering logarithms in precalculus, students build a foundation for calculus, where they will encounter more advanced topics like the natural logarithm and its applications in differentiation and integration.
The logarithmic function is a pivotal concept that shows up in growth models, decay processes, and other real-world contexts. This is why deep comprehension and application in problems, like the one discussed earlier, is critical before advancing further in mathematical studies.
