Problem 15 - Contemporary Precalculus

Question

In Exercises \(11-16,\) let \(u=\ln x\) and \(v=\ln y .\) Write the given expression in terms of u and v. For example, $$\ln x^{3} y=\ln x^{3}+\ln y=3 \ln x+\ln y=3 u+v$$ $$\ln (\sqrt[3]{x^{2} \sqrt{y}})$$

Step-by-Step Solution

Answer

Question: Rewrite the expression \(\ln(\sqrt[3]{x^{2}\sqrt{y}})\) in terms of \(u=\ln x\) and \(v=\ln y\). Solution: \(\ln(\sqrt[3]{x^{2}\sqrt{y}}) = \frac{2}{3}u + \frac{1}{6}v\)

Step 1: Apply the cube root to the expression inside the logarithm

Since we are given \(\ln (\sqrt[3]{x^{2}\sqrt{y}})\), we can write it as: $$ \ln (x^{\frac{2}{3}}y^{\frac{1}{6}})$$ The power of\(x\) becomes \(\frac{2}{3}\) and the power of \(y\) becomes \(\frac{1}{6}\) as we apply the cube root to both.

Step 2: Apply logarithmic properties

Now we apply the properties of logarithm to split the expression. The sum rule states that \(\ln(ab) = \ln a + \ln b\). Also, the power rule states that \(\ln(a^b) = b \ln a\): $$\ln (x^{\frac{2}{3}} y^{\frac{1}{6}}) = \frac{2}{3} \ln x + \frac{1}{6} \ln y$$

Step 3: Rewrite the expression in terms of u and v

Since we are given that \(u=\ln x\) and \(v=\ln y\), we can substitute u and v in the expression: $$\frac{2}{3} \ln x + \frac{1}{6} \ln y = \frac{2}{3} u + \frac{1}{6} v$$ Thus, the final expression in terms of u and v is: $$\ln (\sqrt[3]{x^{2} \sqrt{y}}) = \frac{2}{3}u + \frac{1}{6}v$$

Key Concepts

Logarithm Rules

Logarithms are like the flip side of exponentiation, translating multiplication into addition and division into subtraction. This makes complex multiplication and division problems involving exponents easier to handle.

Core logarithm rules play a crucial role in simplifying logarithmic expressions. The Product Rule states that the logarithm of a product is the sum of the logarithms of the factors: \[\log_b(mn) = \log_b(m) + \log_b(n)\]. The Quotient Rule tells us that the logarithm of a quotient is the difference between the logarithm of the numerator and the logarithm of the denominator: \[\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\]. And the Power Rule simplifies the logarithm of a number raised to a power to the power multiplied by the logarithm of the number: \[\log_b(m^p) = p \cdot \log_b(m)\].

These rules are not just academic; they're practical tools allowing us to untangle complex expressions into manageable pieces. Knowing when and how to apply these rules is crucial for students tackling advanced mathematics.

Exponential Functions

Exponential functions are mathematical expressions where the variable is in the exponent. The general form of an exponential function is \[f(x) = a \times b^x\], where 'a' is the initial value, 'b' is the base or growth factor, and 'x' is the exponent.

These functions are used to model growth and decay in fields ranging from finance to biology. For example, if a bank account earns compound interest, the amount of money grows exponentially over time. In contrast, the radioactive decay of an element decreases exponentially. Moreover, understanding the nature of exponential functions aids in solving practical problems and gives a strong foundation for grasping the logarithmic principles, as logarithms are the inverse operations of exponentials.

Properties of Logarithms

The properties of logarithms are essential for simplifying logarithmic expressions and solving logarithmic equations. Apart from the core rules mentioned previously, other properties include the Change of Base Formula, which allows us to rewrite a logarithm in terms of logarithms of different bases: \[\log_b(x) = \frac{\log_c(x)}{\log_c(b)}\].

This is particularly useful when we need to calculate logarithms on a calculator that only has keys for \(\log\) (base 10) or \(\ln\) (base \(e\)) logarithms. In addition, the Identity Property states that \(\log_b(b) = 1\) because any number raised to the first power is the number itself, and \(\log_b(1) = 0\) because any number raised to the power of zero is one. Hence, understanding these properties can lead to a better grasp of how to manipulate and evaluate logarithmic expressions efficiently and effectively.