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Chapter 12: Problem 70

Use the special properties of logarithms to evaluate each expression. \(\log _{12} 1\)

Short Answer

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Step by step solution

01

Understand the logarithm property

The property of logarithms states that for any base 'b', \(\text{log}_{b}(1) = 0\). This is because any number raised to the power of 0 is 1.
02

Apply the property

Using the property from Step 1, substitute \(\text{log}_{12}(1)\), noting that the base here is 12.
03

Simplify the expression

According to the property, \(\text{log}_{12}(1) = 0\).

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Key Concepts

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

Logarithmic Functions
Logarithmic functions are mathematical operations that are the inverses of exponential functions. When we have an expression like \(\text{log}_{b}(x)\), it is asking the question: 'To what power must the base 'b' be raised to produce 'x'?'. If we say \(\text{log}_{2}(8)\), we are asking: '2 raised to what power equals 8?'. The answer is 3 because \[2^3 = 8.\]
In general, the logarithmic function can be written as: \[\text{log}_{b}(x) = y \rightarrow b^y = x.\][br] This relationship is fundamental to understanding how to manipulate and work with logarithms.
Base and Exponent Relationship
The relationship between base and exponent is crucial in understanding logarithms. In an expression like \(\text{log}_{b}(x)\), the base 'b' must be raised to some exponent 'y' to get 'x'. This can be written as: \[b^y = x\]
For example: \(\text{log}_{10}(100)\) asks '10 raised to what power equals 100?'. The answer is 2, since \[10^2 = 100.\]
This demonstrates how logarithms are used to solve for exponents when the base and the result are known.
Properties of Logarithms
Understanding properties of logarithms makes working with these functions easier. Here are some important ones:
  • Product Property: \(\text{log}_{b}(xy) = \text{log}_{b}(x) + \text{log}_{b}(y)\)
  • Quotient Property: \(\text{log}_{b}(x/y) = \text{log}_{b}(x) - \text{log}_{b}(y)\)
  • Power Property: \(\text{log}_{b}(x^y) = y \text{log}_{b}(x)\)
  • Change of Base Formula: \(\text{log}_{b}(x) = \frac{\text{log}_{k}(x)}{\text{log}_{k}(b)}\). This property allows changing the base of the logarithm to any other base 'k'.
  • Log of 1: \(\text{log}_{b}(1) = 0\). This property is based on the fact that any number raised to the power of 0 is 1.
These properties are powerful tools for simplifying and solving logarithmic expressions. They provide the rules that allow us to manipulate logarithms in a variety of ways.

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Most popular questions from this chapter

$$ \left(\frac{3}{2}\right)^{x}=\frac{16}{81} $$

Use the change-of-base rule (with either common or natural logarithms) to approximate each logarithm to four decimal places. \(\log _{3} \sqrt{2}\)

To four decimal places, the values of \(\log _{10} 2\) and \(\log _{10} 9\) are $$\log _{10} 2=0.3010 \text { and } \log _{10} 9=0.9542$$ Use these values and the properties of logarithms to evaluate each expression. DO NOT USE A CALCULATOR. $$\log _{10} 18$$

Solve each equation. Use natural logarithms. Approximate solutions to three decimal places when appropriate. $$ \ln e^{3 x}=9 $$

Use the properties of logarithms to express each logarithm as a sum or difference of logarithms, or as a single logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{4} \frac{\sqrt[4]{z} \cdot \sqrt[5]{w}}{s^{2}}$$
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