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Exer \(3-4\) : Change to exponential form. (a) \(\log _{3} 81=4\) (b) \(\log _{4} \frac{1}{256}=-4\) (c) \(\log _{r} w=q\) (d) \(\log _{5}(2 x-1)=3\) (e) \(\log _{4} p=5-x\) (f) \(\log _{a} 343=\frac{3}{4}\)

Short Answer

Expert verified
(a) \( 3^4 = 81 \), (b) \( 4^{-4} = \frac{1}{256} \), (c) \( r^q = w \), (d) \( 5^3 = 2x-1 \), (e) \( 4^{5-x} = p \), (f) \( a^{\frac{3}{4}} = 343 \).

Step by step solution

01

Understanding Exponential Form

In logarithm, an equation like \( \log_b a = c \) means that \( b^c = a \). This is because the logarithm asks us the question: 'To what power must the base \( b \) be raised, to yield \( a \)?' Changing from logarithmic to exponential form involves rewriting the equation using this relationship.
02

Convert (a) to Exponential Form

Given \( \log_3 81 = 4 \), apply the principle of exponential form: \( 3^4 = 81 \).
03

Convert (b) to Exponential Form

Given \( \log_4 \frac{1}{256} = -4 \), apply the principle of exponential form: \( 4^{-4} = \frac{1}{256} \).
04

Convert (c) to Exponential Form

Given \( \log_r w = q \), apply the principle of exponential form: \( r^q = w \).
05

Convert (d) to Exponential Form

Given \( \log_5 (2x-1) = 3 \), apply the principle of exponential form: \( 5^3 = 2x - 1 \).
06

Convert (e) to Exponential Form

Given \( \log_4 p = 5-x \), apply the principle of exponential form: \( 4^{5-x} = p \).
07

Convert (f) to Exponential Form

Given \( \log_a 343 = \frac{3}{4} \), apply the principle of exponential form: \( a^{\frac{3}{4}} = 343 \).

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Key Concepts

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

Logarithms
Logarithms are all about unraveling numbers. At their core, they ask the question: "What power should a specific base be raised to in order to reach a particular number?" When you see an expression like \( \log_b a = c \), it is asking: "What power does \( b \) need within to equal \( a \) ?". Understanding this concept is crucial in solving problems involving exponential forms and logarithmic expressions.
Logarithms work inversely compared to exponentiation. It's like understanding the missing piece of a puzzle when trying to determine the number of times you need to multiply the base by itself to reach a certain value. For example, \( \log_2 8 \) yields 3, because \( 2^3 = 8 \).
So, logarithms help in solving problems involving exponential growth or decay, such as calculating compound interest, population growth, or radioactive decay.
Exponential Equations
Exponential equations are where powers and bases play a significant role. These equations typically have a form where a constant base is raised to a variable power. They look like \( b^x = a \), depicting how a base \( b \) is raised to the power \( x \) to produce a result \( a \).
Solving exponential equations often requires rewriting them in a simpler form, which is sometimes achieved by converting them to logarithmic form.
  • Step 1: Isolate the exponential part of the equation. This means getting \( b^x \) by itself.
  • Step 2: Apply logarithms to help unravel the power of the base, making the equation easier to solve.
These equations are key in modeling situations involving continuous change, especially situations involving natural phenomena like populations or financial growth. They are also foundational in dealing with algorithms in computer science and cryptography.
Algebraic Expressions
Algebraic expressions are composed of variables, numbers, and operations. They form the building blocks for algebraic equations and functions.
In expressions like \( 2x - 1 \) or \( 5-x \), the letters represent unknown values or variables. The role of these expressions is to represent relationships and solve problems systematically through logical operations.
When dealing with exponential equations or logarithms, solving the expression typically involves manipulating these variables and constants to find the unknowns. This requires following the systematic application of rules and principles of algebra.
  • Combine like terms.
  • Use distributive property where necessary.
  • Maintain balance by performing identical operations on both sides of an equation.
Mastering algebraic expressions is essential for advancing in algebra and beyond.
Base of Logarithm
The base of a logarithm is the central figure in the logarithmic expression, detailing the number which is being raised to a certain power. In \( \log_b a = c \), \( b \) is the base. It is pivotal because it determines the pathway of transformation in the logarithmic function.
Choosing a base is crucial as it helps define different logarithmic systems, like the common logarithm (base 10) and natural logarithm (base \( e \)). The choice of base impacts how quickly growth or decay is calculated, especially in diverse fields ranging from science to economics.
Each base provides a framework for evaluating and simplifying expressions and is integral in changing equations from logarithmic to exponential form, effectively peeling back the layers of a mathematical problem to reveal deeper insights. Understanding and choosing the right base for logarithms greatly enhances problem-solving capabilities across various disciplines.

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Most popular questions from this chapter

If a savings fund pays interest at a rate of \(3\%\) per year compounded semiannually, how much money invested now will amount to 5000 dollars after 1 year?

Some lending institutions calculate the monthly payment \(M\) on a loan of \(L\) dollars at an interest rate \(r\) (expressed as a decimal) by using the formula $$M=\frac{L r k}{12(k-1)}$$ where \(k=[1+(r / 12)]^{12 t}\) and \(t\) is the number of years that the loan is in effect. Find the largest 25 -year home mortgage that can be obtained at an interest rate of \(7 \%\) if the monthly payment is to be 1500 dollars.

Government spending Federal government expenditures (in billions of dollars) for selected years are listed in the table. $$\begin{array}{|l|c|c|c|c|} \hline \text { Year } & 1910 & 1930 & 1950 & 1970 \\\ \hline \text { Expenditures } & 0.7 & 3.3 & 42.6 & 195.6 \\ \hline \end{array}$$ $$\begin{array}{|l|l|l|l|} \hline \text { Year } & 1980 & 1990 & 2000 \\\ \hline \text { Expenditures } & 590.9 & 1253.1 & 1789.1 \\ \hline \end{array}$$ A. Let \(x=0\) correspond to the year \(1910 .\) Find a function \(A(x)=A_{0} e^{t x}\) that approximates the data, where \(A_{0}\) and \(k\) are constants. Plot the data and \(A\) on the same coordinate axes. B. Use \(A\) to predict graphically the year in which the federal government first spent \(\$ 1\) trillion. (The actual year was 1987 .)

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