Problem 16 - 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, $$\ln x^{3} y=\ln x^{3}+\ln y=3 \ln x+\ln y=3 u+v$$ $$\ln \left(\frac{\sqrt[3]{x^{2} y^{2}}}{x^{5}}\right)$$

Step-by-Step Solution

Answer

Question: Rewrite the expression \(\ln \left(\frac{\sqrt[3]{x^{2} y^{2}}}{x^{5}}\right)\) in terms of u and v, where u = ln(x) and v = ln(y). Answer: The expression can be rewritten as \(-\frac{11}{3}u + \frac{2}{3}v\).

Step 1: Break down the expression into the difference of logarithms using the property ln(a / b) = ln(a) - ln(b)

The given expression is: \(\ln \left(\frac{\sqrt[3]{x^{2} y^{2}}}{x^{5}}\right)\). We'll use the property of logarithms to rewrite this as: $$ \ln(\sqrt[3]{x^{2} y^{2}}) - \ln(x^{5}). $$

Step 2: Rewrite terms in the sum and difference of logarithms using the properties ln(a^b) = b * ln(a) and ln(a*b) = ln(a) + ln(b)

Now, let's rewrite the two terms separately: $$ \begin{aligned} \ln(\sqrt[3]{x^{2} y^{2}}) &= \frac{1}{3}\ln(x^{2} y^{2}), \quad\text{by property ln(a^b) = b * ln(a)} \\ &= \frac{1}{3}(\ln(x^{2}) + \ln(y^{2})), \quad\text{by property ln(a * b) = ln(a) + ln(b)} \\ &= \frac{1}{3}(2\ln(x) + 2\ln(y)), \quad\text{by property ln(a^b) = b * ln(a)} \\ &= \frac{2}{3}(\ln(x) + \ln(y)). \end{aligned} $$ Next term is: $$ \begin{aligned} \ln(x^{5}) = 5\ln(x), \quad\text{by property ln(a^b) = b * ln(a)} \end{aligned} $$

Step 3: Substitute u and v in the terms we found in step 2

Now, recall that u = ln(x) and v = ln(y). We'll substitute these into the two terms we found in step 2: $$ \frac{2}{3}(\ln(x) + \ln(y)) = \frac{2}{3}(u + v). $$ $$ 5\ln(x) = 5u. $$

Step 4: Combine terms and write the final expression in terms of u and v

Finally, we'll combine the terms: $$ \ln \left(\frac{\sqrt[3]{x^{2} y^{2}}}{x^{5}}\right) = \frac{2}{3}(u + v) - 5u. $$ So the expression is: $$ \frac{2}{3}u + \frac{2}{3}v - 5u = \boxed{-\frac{11}{3}u + \frac{2}{3}v}. $$

Key Concepts

Natural Logarithm Properties

Understanding natural logarithm properties is essential for simplifying logarithmic expressions. The natural logarithm, denoted as \( \ln(x) \), has a base of \( e \), where \( e \) is the irrational Euler's number, approximately 2.71828.

In the exercise, we encounter two critical properties:

• \( \ln(a^b) = b \cdot \ln(a) \), which tells us that a logarithm of a power can be rewritten as the exponent times the logarithm of the base.
• \( \ln(a \times b) = \ln(a) + \ln(b) \), which represents the logarithm of a product as the sum of logarithms.

These properties enable the algebraic manipulation of the logarithmic expression to simplify it into terms of \( u \) and \( v \), which are easier to handle.

Logarithm Rules

Grasping the fundamental logarithm rules will allow you to combine and simplify logarithmic expressions with confidence. Besides the properties used in the natural logarithm, additional logarithm rules include:

• The logarithm of a quotient: \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \), which we saw in action in the given exercise when breaking down the initial expression into the difference of two logarithms.
• The logarithm of one: \( \ln(1) = 0 \), which is handy in eliminating terms.
• The change-of-base formula, which wasn't directly used in this exercise, but is important when you are dealing with logarithms of bases other than \( e \).

Once you become familiar with these rules, you'll find that intricate logarithmic expressions can often be deconstructed and reassembled into simpler forms.

Algebraic Manipulation

Algebraic manipulation is the art of rearranging, combining, and simplifying expressions to solve equations or make them easier to understand. In the context of logarithms, this often involves using the properties and rules of logarithms discussed above.

Applying Algebraic Manipulation

In our exercise, algebraic manipulation is used to break down the logarithm of a root and express it in terms of simple logarithms of \( x \) and \( y \) before substituting \( u \) and \( v \). Notice how the manipulation allowed us to transition from a complex logarithmic expression to a much simpler one using the defined variables. Mastering algebraic manipulation is not only about knowing the rules but also understanding when and how to apply them efficiently, often making a seemingly daunting task manageable.

By becoming proficient in algebraic manipulation, you can approach a wide array of mathematical problems with a toolkit equipped to simplify and solve them systematically.