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Express the quantity in terms of natural logarithms. $$\log _{10} e$$

Short Answer

Expert verified
\( \log_{10} e = \frac{1}{\ln 10} \)

Step by step solution

01

Understand the Conversion from Base 10 to Base e

We are provided with the expression \( \log_{10} e \), which is a logarithm with base 10. Our task is to express this in terms of natural logarithms, which are logarithms with base \( e \). We need to use the change of base formula.
02

Apply the Change of Base Formula

The change of base formula states that for any positive numbers \( a \), \( b \), and \( c \), where \( a eq 1 \), the expression \( \log_a b \) can be rewritten in terms of natural logarithms as \[ \log_a b = \frac{\ln b}{\ln a} \].
03

Substitute into the Change of Base Formula

Using \( b = e \) and \( a = 10 \), substitute these values into the formula: \[ \log_{10} e = \frac{\ln e}{\ln 10} \].
04

Simplify the Expression

We know that \( \ln e = 1 \) because the natural logarithm of \( e \) is 1. Hence, replace \( \ln e \) with 1: \[ \log_{10} e = \frac{1}{\ln 10} \].

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Key Concepts

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

Change of Base Formula
The change of base formula is an essential tool for converting logarithms from one base to another. This is especially useful when you want to express logarithmic expressions using a base that is more convenient for your calculations, such as the natural base, denoted by the number \( e \).

Here's how the change of base formula works: If you have a logarithm \( \log_a b \), you can convert it into a logarithm with a different base \( c \) using the equation:
  • \( \log_a b = \frac{\log_c b}{\log_c a} \)
For natural logarithms, which use base \( e \), \( \log_a b \) can simply be written as:
  • \( \log_a b = \frac{\ln b}{\ln a} \)
This capability allows us to conveniently use calculators that may only provide natural logarithm functions, or to solve mathematical problems that require a specific base like \( e \). By applying this formula, complex base conversions become manageable, and computations involving logarithms remain consistent and practical.
Logarithmic Expressions
Logarithmic expressions involve the use of logarithm functions, which are the inverse operations of exponentials. Here's a simple breakdown of how logarithms work:

  • \( \log_b a \) is the power to which the base \( b \) must be raised to obtain the number \( a \).
  • For instance, \( \log_{10} 100 = 2 \) because \( 10^2 = 100 \).
  • Natural logarithms, denoted as \( \ln \), use the base \( e \) where \( e \approx 2.718 \).
Logarithmic expressions are not limited to just whole numbers. They can include fractions and decimals, and are frequently used in areas like science and engineering, especially when working with exponential growth, decay, and pH calculations. Understanding how to manipulate and convert these expressions is crucial for solving various mathematical problems efficiently. The ability to express logarithmic terms in different bases is a vital skill, helping to simplify and solve equations that would otherwise be difficult to manage.
Base Conversion
Base conversion in the realm of logarithms refers to changing a logarithm's base, often to a base that simplifies calculations, such as converting a common logarithm (base 10) into a natural logarithm (base \( e \)).

In our example, \( \log_{10} e \), the task is to express it in terms of natural logarithms. To achieve this, we use the change of base formula:
  • First, identify the original base (10) and the target base (\( e \)).
  • Replace the original expression using the formula: \( \log_{10} e = \frac{\ln e}{\ln 10} \).
  • Simplify the result using known values, such as \( \ln e = 1 \).
  • This simplifies to \( \frac{1}{\ln 10} \).
By converting the base, calculations become more straightforward, especially when using calculators. It allows consistency in mathematical operations, no matter the initial base. Base conversion is instrumental in bringing flexibility and simplicity to handling various mathematical tasks, making it easier to solve and interpret logarithmic expressions.

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Most popular questions from this chapter

The Chernobyl nuclear explosion (in the former Soviet Union, on April 26,1986 ) released large amounts of radioactive substances into the atmosphere. These substances included cesium-137, iodine-131, and strontium-90. Although the radioactive material covered many countries, the actual amount and intensity of the fallout varied greatly from country to country, due to vagaries of the weather and the winds. One area that was particularly hard hit was Lapland, where heavy rainfall occurred just when the Chernobyl cloud was overhead. (a) Many of the pastures in Lapland were contaminated with cesium-137, a radioactive substance with a half- life of 33 years. If the amount of cesium- 137 was found to be ten times the normal level, how long would it take until the level returned to normal? Hint: Let \(\mathcal{N}_{0}\) be the amount that is ten times the normal level. Then you want to find the time when \(\mathcal{N}(t)=\mathcal{N}_{0} / 10\) (b) Follow part (a), but assume that the amount of cesium-137 was 100 times the normal level. Remark: Several days after the explosion, it was reported that the level of cesium- 137 in the air over Sweden was 10,000 times the normal level. Fortunately there was little or no rainfall.

Solve each equation and solve for \(x\) in terms of the other letters. $$\beta=10 \log _{10}\left(x / x_{0}\right)$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$e^{1 /(x-1)}>1$$

The following extract is from an article by Kim Murphy that appeared in the Los Angeles Times on September 14 1994 CAIRO-Over a chorus of reservations from Latin America and Islamic countries still troubled about abortion and family issues, nearly 180 nations adopted a wide-ranging plan Tuesday on global population, the first in history to obtain partial endorsement from the Vatican. The plan, approved on the final day of the U.N. population conference here, for the first time tries to limit the growth of the world's population by preventing it from exceeding 7.2 billion people over the next two decades. (a) In 1995 the world population was 5.7 billion, with a relative growth rate of \(1.6 \% /\) year. Assuming continued exponential growth at this rate, make a projection for the world population in the year \(2020 .\) Round off the answer to one decimal place. How does your answer compare to the target value of 7.2 billion mentioned in the article? (b) As in part (a), assume that in 1995 the world population was 5.7 billion. Determine a value for the growth constant \(k\) so that exponential growth throughout the years \(1995-2020\) leads to a world population of 7.2 billion in the year 2020 .

In the text we showed that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is constant for all time intervals of unit length, \([t, t+1] .\) Recall that we did this by computing the relative change \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)\) and noting that the result was a constant, independent of \(t .\) (If you've completed the previous exercise, you've done this calculation for yourself.) Now consider a time interval of arbitrary length, \([t, t+d] .\) The relative change in the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) over this time interval is \([\mathcal{N}(t+d)-\mathcal{N}(t)] / \mathcal{N}(t) .\) Show that this quantity is a constant, independent of \(t .\) (The expression that you obtain for the constant will contain \(e\) and \(d\), but not \(t\) As a check on your work, replace \(d\) by 1 in the expression you obtain and make sure the result is the same as that in the text where we worked with intervals of length \(d=1 .)\)

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