Problem 2 - Thomas Calculus

Question

Express the following logarithms in terms of \(\ln 5\) and \(\ln 7\) $$\begin{array}{lll}{\text { a. } \ln (1 / 125)} & {\text { b. } \ln 9.8} & {\text { c. } \ln 7 \sqrt{7}} \\ {\text { d. } \ln 1225} & {\text { e. } \ln 0.056} \\ {\text { f. }} {(\ln 35+\ln (1 / 7)) /(\ln 25)}\end{array}$$

Step-by-Step Solution

Answer

a. \(-3\ln 5\), b. \(2\ln 7 - \ln 5\), c. \(\frac{3}{2}\ln 7\), d. \(2\ln 5 + 2\ln 7\), e. \(\ln 7 + 3\ln 2 - 3\ln 10\), f. \(\frac{1}{2}\).

Step 1: Solve for \(\ln (1/125)\)

Recall that \(125 = 5^3\). Using the logarithm property \(\ln (a/b) = \ln a - \ln b\), then:\[\ln \left(\frac{1}{125}\right) = \ln 1 - \ln 125 = -\ln 125.\]Also, \(\ln 125 = \ln (5^3) = 3 \ln 5\). Therefore, the expression simplifies to:\[ \ln \left(\frac{1}{125}\right) = -3 \ln 5.\]

Step 2: Solve for \(\ln 9.8\)

Observe that \(9.8 = 49/5 = 7^2/5\). Apply logarithm properties:\[\ln 9.8 = \ln \left(\frac{7^2}{5}\right) = \ln(7^2) - \ln 5 = 2\ln 7 - \ln 5.\]

Step 3: Solve for \(\ln 7 \sqrt{7}\)

The expression \(7 \sqrt{7}\) can be rewritten as \(7^{3/2}\). Using the property \(\ln a^b = b \ln a\), we have:\[\ln (7 \sqrt{7}) = \ln (7^{3/2}) = \frac{3}{2} \ln 7.\]

Step 4: Solve for \(\ln 1225\)

Note that \(1225 = 35^2 = (5 \cdot 7)^2\). Applying properties of logarithms:\[\ln 1225 = \ln ((5 \cdot 7)^2) = 2(\ln 5 + \ln 7) = 2 \ln 5 + 2 \ln 7.\]

Step 5: Solve for \(\ln 0.056\)

Recognize \(0.056 = 5.6 \times 10^{-2} = \frac{56}{1000}\). Thus:\[\ln 0.056 = \ln \left(\frac{56}{1000}\right) = \ln 56 - \ln 1000 = \ln (7 \cdot 8) - 3 \ln 10\]Involving purely natural logarithms:\[ \ln 56 = \ln (7 \cdot 8) = \ln 7 + \ln 8 = \ln 7 + \ln (2^3) = \ln 7 + 3 \ln 2\]Therefore:\[\ln 0.056 = \ln 7 + 3 \ln 2 - 3 \ln 10\]

Step 6: Solve for \((\ln 35 + \ln (1/7)) / \ln 25\)

First, simplify the numerator:\[\ln 35 + \ln \left(\frac{1}{7}\right) = \ln (5 \cdot 7) - \ln 7 = \ln 5.\]Now, for the denominator:\[\ln 25 = \ln (5^2) = 2 \ln 5.\]Thus, the entire expression becomes:\[\frac{\ln 35 + \ln (1/7)}{\ln 25} = \frac{\ln 5}{2 \ln 5} = \frac{1}{2}.\]

Key Concepts

Natural Logarithms

Natural logarithms, often denoted as \(\ln\), are a special type of logarithm with the base \(e\), where \(e\) is the mathematical constant approximately equal to 2.71828. They are widely used in various fields like mathematics, science, and engineering because they naturally occur in many growth processes. For example, in the original exercise, each expression is converted into a natural logarithm form using the properties of logarithms, which allows us to conveniently express complex relationships in terms of simpler base elements like \(\ln 5\) and \(\ln 7\). These conversions often involve a deeper understanding of mathematical expressions and the relationships between their components.

Logarithm Properties

Logarithmic properties are key tools when dealing with logarithms. These properties help simplify and manipulate logarithmic expressions.

• Product Property: \(\ln(a \cdot b) = \ln a + \ln b\)
• Quotient Property: \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\)
• Power Property: \(\ln(a^b) = b \cdot \ln a\)

These properties are used repeatedly in the exercise to break down complex expressions into more manageable parts. For instance, when calculating \(\ln(1/125)\), the quotient property is used to express the logarithm in subtraction form, and the power property simplifies \(\ln(5^3)\) into \(3 \ln 5\). Understanding these properties is crucial for simplifying and solving logarithmic expressions efficiently.

Simplifying Logarithms

Simplifying logarithmic expressions involves using logarithm properties to rewrite expressions into simpler or more useful forms. This often includes converting products, quotients, or powers within a logarithmic statement into basic sums or differences that are easier to handle. As seen in the exercise:

• \(\ln 9.8\) was rewritten by recognizing it as \(\frac{7^2}{5}\), thus sharing it into \(2\ln 7 - \ln 5\).
• \(\ln 7 \sqrt{7}\) was simplified to \(\frac{3}{2}\ln 7\), showcasing the use of exponent forms for simpler computation.
• These steps demonstrate algebraic manipulation, breaking down more complex forms into simpler, base logarithms.

This technique is central to not only make expressions easier to work with but also to understand their intrinsic relationships.

Algebraic Manipulation

Algebraic manipulation in logarithmic contexts refers to the strategic use of algebra to rearrange and simplify expressions. It requires a good grasp of both algebra and the properties of logarithms. For example:

• In the problem \((\ln 35 + \ln (1/7)) / \ln 25\), the use of algebraic manipulation simplifies it to \(\frac{1}{2}\) by applying the properties of products and powers on logarithms.
• Breaking down \(\ln 1225\) involved recognizing common factors and simplifying them using known logarithmic identities.

Such manipulation not only helps in deriving exact answers but also assists in gaining deeper insight into the behavior of logarithmic functions. It shows how expressions can be transformed into equivalent forms that are easier to interpret or compute.