/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 In Exercises \(7-14,\) use \(\lo... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercises \(7-14,\) use \(\log 2 \approx 0.3010, \log 5 \approx 0.6990,\) and \(\log 7 \approx 0.8451\) to evaluate each logarithm without using a calculator. Then check your answer using a calculator. $$\log \frac{2}{5}$$

Short Answer

Expert verified
Therefore, the evaluation of the given expression \(\log \frac{2}{5}\) without using a calculator is \(-0.3980\).

Step by step solution

01

Identifying the Given Logarithm

\(\log \left(\frac{2}{5}\right)\) is the given logarithm. It can be broken down into a subtraction of two logs, based on the logarithm property \(\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\). Hence, this is equal to \(\log(2) - \log(5)\).
02

Substituting the Values

Replacing the values of \(\log(2)\) and \(\log(5)\) with the approximate given values, we get \(0.3010 - 0.6990\).
03

Perform the Operation

Subtract the values to get the final answer. The answer to \(0.3010 - 0.6990\) is \(-0.3980\).

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Key Concepts

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

Logarithmic Properties
Logarithmic properties are rules we use to transform, simplify, or solve equations involving logarithms. These rules make it easier to handle complex logarithmic expressions. One key property used in this exercise is the quotient rule for logarithms:
\[ \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \]
This formula helps split a logarithm of a division into the subtraction of two logarithms. It can simplify calculations and allow us to solve problems by hand without a calculator.

Other properties also help in manipulating and evaluating logarithmic expressions:
  • The product rule: \( \log(ab) = \log(a) + \log(b) \)
  • The power rule: \( \log(a^n) = n \cdot \log(a) \)
  • The change of base formula: \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), where \( c \) is any positive number
Understanding these properties allows you to approach logarithmic problems with confidence. You can simplify and reorganize expressions to make problem-solving more intuitive.
Subtraction of Logarithms
Subtraction of logarithms comes up frequently, especially when using the quotient rule. If you have a logarithm of a fraction, like \( \log\left(\frac{2}{5}\right) \), subtraction is a handy tool for simplification. By applying the rule:
\[ \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \]we convert a division within a logarithm into the subtraction of log values.

Our exercise demonstrates this neatly. We can evaluate \( \log\left(\frac{2}{5}\right) \) by finding \( \log(2) \) and \( \log(5) \) separately and then subtracting the latter from the former:
  • \( \log(2) \approx 0.3010 \)
  • \( \log(5) \approx 0.6990 \)
    The calculation then becomes:
  • \( 0.3010 - 0.6990 = -0.3980 \)
The subtraction allows us to solve the problem efficiently and understand the transformation of logarithmic expressions more deeply.
Approximations in Logarithms
Approximations are crucial in dealing with logarithms when precise values aren't available or necessary. We often approximate logarithmic values to render calculations more manageable without advanced tools like calculators.

In this exercise, we used approximations:
  • \( \log(2) \approx 0.3010 \)
  • \( \log(5) \approx 0.6990 \)
  • \( \log(7) \approx 0.8451 \)
These values aren't precise, but they are close enough to provide meaningful results.

When checking work, confirm with a calculator or exact values when possible to ensure your approximated result closely resembles the true value. Approximations enable practical, on-the-spot computations, especially in educational settings where understanding the method is as vital as the precision of the result.

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Most popular questions from this chapter

Use the change-of-base formula to evaluate each logarithm using a calculator. Round answers to four decimal places. $$. \log _{7} 150$$

Applications In this set of exercises, you will use inverse functions to study real-world problems. A woman's dress size in the United States can be converted to a woman's dress size in France by using the function \(f(s)=s+30,\) where \(s\) takes on all even values from 2 to \(24,\) inclusive. (Source: www.onlineconversion \(. \operatorname{com})\) (a) What is the range of \(f ?\) (b) Find the inverse of \(f\) and interpret it.

Find the inverse of the given function. Then graph the given function and its inverse on the same set of axes. $$g(x)=2 x^{5}-6$$

Consider the two functions \(f(x)=2 x\) and \(g(x)=2^{x}.\) (a) Make a table of values for \(f(x)\) and \(g(x),\) with \(x\) ranging from -1 to 4 in steps of 0.5. (b) Find the interval(s) on which \(2 x<2^{x}.\) (c) Find the interval(s) on which \(2 x>2^{x}.\) (d) Using your table from part (a) as an aid, state what happens to the value of \(f(x)\) if \(x\) is increased by 1 unit. (e) Using your table from part (a) as an aid, state what happens to the value of \(g(x)\) if \(x\) is increased by 1 unit. (f) Using your answers from parts (c) and (d) as an aid, explain why the value of \(g(x)\) is increasing much faster than the value of \(f(x).\)

A new car that costs $$\$ 25,000$$ depreciates to $$80 \%$$ of its value in 3 years. (a) Assume the depreciation is linear. What is the linear function that models the value of this car \(t\) years after purchase? (b) Assume the value of the car is given by an exponential function \(y=A e^{h t},\) where \(A\) is the initial price of the car. Find the value of the constant \(k\) and the exponential function. (c) Using the linear model found in part (a), find the value of the car 5 years after purchase. Do the same using the exponential model found in part (b). (d) Graph both models using a graphing utility. Which model do you think is more realistic, and why?

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