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Chapter 11: Problem 3

Rewrite the equation using exponents instead of logarithms. $$ \log 20=1.301 $$

Short Answer

Expert verified
Question: Rewrite the logarithmic equation $\log 20 = 1.301$ using exponents. Answer: $10^{1.301} = 20$

Step by step solution

01

Identify the base of the logarithm

In this case, since no base is mentioned explicitly, we consider the base to be 10 (common logarithm).
02

Convert to exponential form

Since the base is 10, we can rewrite the given equation in exponential form as: $$ 10^{1.301} = 20 $$ This equation states that 10 raised to the power of 1.301 equals 20, which is the given equation expressed using exponents instead of logarithms.

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Key Concepts

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

Logarithmic Functions
Logarithmic functions play an essential role in mathematics, especially when dealing with exponents. Simply put, a logarithm answers the question: "To what exponent must we raise the base to get a certain number?" For example, if you have \( \log_b(x) = y \), it means \( b^y = x \). This relationship is central to understanding how logarithms function.
  • They act as the inverse operation of exponentiation.
  • They're often used to solve equations where the variable is an exponent.
Logarithms have applications in various real-world scenarios, such as in calculating interest rates, sound levels (decibels), and even in scientific calculations where very large or small numbers are involved. By mastering logarithmic functions, you equip yourself with the tools to tackle a wide range of mathematical problems.
Base of a Logarithm
The base of a logarithm is crucial because it determines the logarithm's scale and behavior. The base is the number that is raised to the power of the logarithm result. For instance, in the logarithm \( \log_b(x) = y \), \( b \) is the base. The base specifies the multiplicative rate at which the quantities upset by the logarithm increase.
  • Common bases you might encounter include 10 (decimal system), 2 (binary system), and \( e \) (natural logarithms).
  • The choice of base affects the result and interpretation of a logarithm.
When a base isn't explicitly stated, as in the problem given, it's often assumed to be 10 (called a common logarithm). Understanding the base helps translate logarithmic expressions into a more intuitive form.
Exponential Form
Converting between logarithmic and exponential forms is a skill that simplifies solving mathematical equations. Exponential form expresses a number as a base raised to a power (exponent). For example, \( b^y = x \) means the same as \( \log_b(x) = y \).
  • This conversion is particularly useful for solving equations that involve logarithms or exponents.
  • Knowing the exponential form allows you to interpret logarithmic equations more intuitively.
In our example, the equation \( \log 20=1.301 \) was converted to \( 10^{1.301} = 20 \). Here, 10 is the base, 1.301 is the exponent, and the result is 20. Recognizing how to move fluidly between logarithmic and exponential forms makes it easier to solve related problems.
Common Logarithms
Common logarithms are logarithms with base 10. They are frequently used in calculations because our number system is based on powers of 10. Common logarithms are indicated simply by \( \log(x) \) without explicitly mentioning the base.
  • They simplify operations on numbers that are vastly different in scale.
  • These are often used in scientific contexts and are standard in many calculators.
In the problem, \( \log 20 = 1.301 \) is a common logarithm, meaning the base is 10. By converting this into the exponential form \( 10^{1.301} = 20 \), you're able to see how a seemingly abstract logarithm has a very concrete interpretation in terms of exponents.

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Most popular questions from this chapter

Solve the equations, first approximately, as in Example \(1,\) by filling in the given table, and th猫n to four decimal places by using logarithms. $$ \begin{aligned} &\text { Table } 11.3 \text { Solve } 10^{x}=500\\\ &\begin{array}{c|c|c|c|c} \hline x & 2.6 & 2.7 & 2.8 & 2.9 \\ \hline 10^{x} & & & & \\ \hline \end{array} \end{aligned} $$

A radioactive substance decays at a constant percentage rate per year. (a) Find the half-life if it decays at a rate of (i) \(10 \%\) per year. (ii) \(19 \%\) per year. (b) Compare your answers in parts (i) and (ii). Why is one exactly half the other?

Evaluate the expressions without using a calculator. Verify your answers. Example. We have \(\log _{2} 32=5\) because \(2^{5}=32\). $$ \log _{20} 400 $$

Evaluate the expressions without using a calculator. Verify your answers. Example. We have \(\log _{2} 32=5\) because \(2^{5}=32\). $$ \log _{9} 9 $$

The size of two towns \(t\) years after 2000 is given by \(u(t)=1200(1.019)^{t}\) and \(v(t)=1550(1.038)^{t} .\) Solve the equation \(u(t)=v(t) .\) What does the solution tell you about the towns?
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