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Chapter 4: Problem 18

Exer. 17-20: Find the number. (a) \(10^{\log 7}\) (b) \(\log 10^{-6}\) (c) \(\log 100,000\) (d) \(\log 0.001\) (e) \(10^{-1+\log 5}\)

Short Answer

Expert verified
(a) 7, (b) -6, (c) 5, (d) -3, (e) 0.5.

Step by step solution

01

Solve (a)

The expression given is \(10^{\log 7}\). Using the property \(a^{\log_a b} = b\), we can directly find that \(10^{\log 7} = 7\).
02

Solve (b)

The expression is \(\log 10^{-6}\). We use the property of logarithms that states \(\log 10^a = a\). Here, \(a = -6\). Therefore \(\log 10^{-6} = -6\).
03

Solve (c)

The expression is \(\log 100,000\). Recognize that \(100,000 = 10^5\). Using \(\log 10^a = a\), the answer is \(\log 100,000 = 5\).
04

Solve (d)

The expression is \(\log 0.001\). Note that \(0.001 = 10^{-3}\). Applying the property \(\log 10^a = a\), we find \(\log 0.001 = -3\).
05

Solve (e)

The expression is \(10^{-1+\log 5}\). Using the properties \(10^{-1} = \frac{1}{10}\) and \(10^{\log 5} = 5\), we apply the identity \(10^a \, 10^b = 10^{a+b}\) to simplify: \(10^{-1+\log 5} = 10^{-1} \times 10^{\log 5} = \frac{1}{10} \times 5 = 0.5\).

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Key Concepts

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

Properties of Logarithms
Logarithms have powerful properties that help simplify complex equations. The key properties include:
  • Product Rule: \(\log_b(xy) = \log_b x + \log_b y\). This states that the logarithm of a product is the sum of the logarithms of the factors.
  • Quotient Rule: \(\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y\). This indicates that the logarithm of a quotient is the difference of the logarithms.
  • Power Rule: \(\log_b(x^a) = a \cdot \log_b x\). This tells us that the logarithm of an exponentiated term is the power times the logarithm of the base.
  • Change of Base Formula: \(\log_b x = \frac{\log_k x}{\log_k b}\). This formula helps translate logs from one base to another.
  • Exponential Property: \(a^{\log_a x} = x\). This means if you raise a base to the log of the base, you're left with the argument.
Understanding these properties is crucial since they allow us to manipulate and solve logarithmic expressions more flexibly.
Base 10
Logarithms often use base 10, also called the 'common logarithm'. When you see \(\log x\), it's typically understood to mean \(\log_{10} x\). Base 10 logarithms are prevalent in various fields due to the decimal system, making them relatable and easy to calculate.
Some unique aspects of base 10 include:
  • Easy to Calculate: The log scale at base 10 simplifies many practical problems, especially when dealing with orders of magnitude.
  • Used in Scientific Calculators: Most scientific calculators have a dedicated button for base 10 logarithms, reflecting their wide usage.
  • Historical Relevance: Before electronic calculators, logarithm tables based on base 10 were used to facilitate computation.
Base 10 helps to find exponential relationships easily, such as determining growth rates or sound intensity levels.
Exponentiation
Exponentiation involves raising a base number to a certain power. In the context of logarithms, it ties directly into their properties.
Here are important insights about exponentiation:
  • Inverse of Logarithms: If \(\log_b(x) = y\), then \(b^y = x\), showing the intrinsic dual relationship.
  • Key Property: In exercises like \(10^{\log 7}\), exponentiation and logarithms cancel each other when bases match, simplifying to just 7.
  • Mathematical Flexibility: Exponents can handle small increments of scaling, making them versatile across sciences.
When numbers integrate both exponents and logs, solutions can sometimes be intuitive and faster, hinging on their reciprocal nature.
Logarithmic Equations
When working with logarithmic equations, you're dealing with equations that involve unknowns within logarithm expressions. Solving them often requires understanding and applying logarithm properties.
  • Isolation of the Logarithmic Term: First step is to isolate the log term as was seen in the exercise \(\log 10^{-6} \), where simplifying requires applying the exponent directly.
  • Exponentiate Both Sides: To solve for the unknown, you typically exponentiate both sides of an equation to eliminate the log.
  • Utilizing Logarithm Properties: Often involves expressing terms in such a way to match logarithmic forms for easy computation.
Building a solid foundation in logarithmic properties and exponentiation is essential to tackle more complex and practical problems, making these equations insightful in various scientific fields.

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Most popular questions from this chapter

Exer. 83-84: Approximate the function at the value of \(x\) to four decimal places. (a) \(f(x)=\log \left(2 x^{2}+1\right)-10^{-x}, \quad x=1.95\) (b) \(g(x)=\frac{x-3.4}{\ln x+4}, \quad x=0.55\)

Real estate appreciation If the value of real estate increases at a rate of 4\% per year, after \(t\) years the value \(V\) of a house purchased for \(P\) dollars is \(V=P(1.04)^{t} .\) A graph for the value of a house purchased for 80,000 dollars in 1986 is shown in the figure. Approximate the value of the house, to the nearest 1000 dollars, in the year 2016 . (GRAPH CAN'T COPY)

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. (a) Find the monthly payment on a 30 -year 250,000 dollars home mortgage if the interest rate is \(8 \%\) (b) Find the total interest paid on the loan in part (a).

Exer. \(77-78:\) Graph \(f\) and \(g\) on the same coordinate plane, and estimate the solution of the inequality \(f(x)>g(x)\) $$f(x)=3^{-x}-4^{0.2 x} ; \quad g(x)=\ln (1.2)-x$$

In \(1974,\) Johnny Miller won 8 tournaments on the PGA tour and accumulated 353,022 dollars in official season earnings. In \(1999,\) Tiger Woods accumulated 6,616,585 dollars with a similar record. (a) Suppose the monthly inflation rate from 1974 to 1999 was \(0.0025(3 \% / \mathrm{yr}) .\) Use the compound interest formula to estimate the equivalent value of Miller's winnings in the year \(1999 .\) Compare your answer with that from an inflation calculation on the web (e.g., bls.gov/cpi/home.htm). (b) Find the annual interest rate needed for Miller's winnings to be equivalent in value to Woods's winnings. (c) What type of function did you use in part (a)? part (b)?
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