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Chapter 5: Problem 4

Simplify the expression by using the definition and properties of logarithms. $$\log _{9} 25-\log _{9} 75$$

Short Answer

Expert verified
The simplified expression is \( \log_{9} \left( \frac{1}{3} \right) \).

Step by step solution

01

Identify the Logarithm Property

Recognize that the expression \( \log_{9} 25 - \log_{9} 75 \) can be simplified using the properties of logarithms. Specifically, use the property \( \log_{b} M - \log_{b} N = \log_{b} \left( \frac{M}{N} \right) \).
02

Apply the Logarithm Property

Apply the property from Step 1 to simplify the expression: \( \log_{9} 25 - \log_{9} 75 = \log_{9} \left( \frac{25}{75} \right) \).
03

Simplify the Fraction

Simplify the fraction inside the logarithm. The expression becomes \( \log_{9} \left( \frac{1}{3} \right) \) because \( \frac{25}{75} = \frac{1}{3} \).
04

Simplified Expression

Conclude that the simplified expression is \( \log_{9} \left( \frac{1}{3} \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithm Properties
Understanding the properties of logarithms is essential when simplifying expressions. Logarithms are exponents and they help express numbers in a different base. Here are few key properties:
  • Product Property: This states that the logarithm of a product is the sum of the logarithms. Mathematically, it's expressed as: \( \log_b(M \cdot N) = \log_b(M) + \log_b(N) \).
  • Quotient Property: This property is used to simplify the logarithm of a quotient. It tells us that \( \log_b \left( \frac{M}{N} \right) = \log_b(M) - \log_b(N) \).
  • Power Property: This property helps when dealing with powers, expressed as: \( \log_b(M^p) = p \cdot \log_b(M) \).
These properties allow us to manipulate logarithmic expressions, making them easier to solve. In the example given, the quotient property is key, converting subtraction of two logarithms into a single logarithm of a division.
Simplification of Expressions
Simplifying expressions in mathematics is about making them easier to understand or work with. In logarithmic expressions, we look to decrease the complexity by using properties as seen in the original step-by-step solution.
  • Recognizing applicable properties: When presented with a logarithmic expression like \( \log_{9} 25 - \log_{9} 75 \), identify properties that reduce multiple terms into one.
  • Reduction of fractions: After applying the property, simplify the underlying fraction, as simplifying \( \frac{25}{75} \) to \( \frac{1}{3} \) was demonstrated.
  • Rewriting into simplest base: Once reduced, expressions become straightforward, such as simplifying to \( \log_{9} \left( \frac{1}{3} \right) \).
Simplification is about clarity. The simpler the expression, the more manageable it becomes for further analysis or computation.
Precalculus Concepts
Precalculus serves as a bridge between arithmetic, algebra, and calculus. It introduces concepts that are not only academic but also applicable in real-world problems.
  • Understanding functions: Precalculus considers various functions, including logarithmic functions, which describe a wide range of phenomena.
  • Analysis of expressions: Understanding how to manipulate and simplify expressions using properties is foundational in precalculus. It prepares students for more advanced calculus challenges.
  • Real-world application: Logarithms play a role in fields such as computer science, financial modeling, and natural sciences, where exponential growth and decay problems often arise.
This foundation is vital for students progressing to calculus, enabling them to handle more complex mathematical modeling efficiently. Understanding these precalculus concepts ensures a smoother transition into higher-level mathematics.

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Most popular questions from this chapter

The half-life of strontium- 90 is 28 years. How much of a \(10-\mathrm{g}\) sample will remain after (a) 1 year? (b) 10 years?

The radioactive isotope sodium-24 is used as a tracer to measure the rate of flow in an artery or vein. The half-life of sodium-24 is 14.9 hours. Suppose that a hospital buys a \(40-\mathrm{g}\) sample of sodium- 24 (a) How much of the sample will remain after 48 hours? (b) How long will it be until only 1 gram remains?

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$e^{(1 / x)-1}>1$$

(a) Let \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t} .\) Show that \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)=e^{k}-1 .\) (This is actually done in detail in the text. So, ideally, you should look back only if you get stuck or want to check your answer.) (b) Assume as given the following approximation, which was introduced in Exercise 26 of Section 5.2 \(e^{x} \approx x+1 \quad\) provided \(x\) is close to zero Use this approximation to explain why \(e^{k}-1 \approx k\) provided that \(k\) is close to zero. Remark: Combining this result with that in part (a), we conclude that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is approximately equal to the growth constant \(k .\) As explained in the text, this is one of the reasons why in applications we've not distinguished between the relative growth rate and the decay constant \(k\)

Use the half-life information to complete each table. (The formula \(\mathcal{N}=\mathcal{N}_{0} e^{k t}\) is not required.) (a) Uranium-228: half-life \(=550\) seconds$$\begin{array}{llllll}t \text { (seconds) } & 0 & 550 & 1100 & 1650 & 2200 \\\\\mathcal{N} \text { (grams) } & 8 & & & \\\\\hline\end{array}$$ (b) Uranium-238: half-life \(=4.9 \times 10^{9}\) years$$\begin{array}{lcccccc}\hline \multirow{2}{*}\begin{array}{l}\text { t (years) } \\\\\mathcal{N} \text { (grams) }\end{array} & \multicolumn{2}{c}0 \\\& \multicolumn{2}{c}10 & 5 & 2.5 & 1.25 & 0.625 \\\\\hline\end{array}$$
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