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Chapter 5: Problem 53

Express the quantity in terms of base 10 logarithms. $$\ln 3$$

Short Answer

Expert verified
\( \ln 3 = \log_{10} 3 \cdot \ln 10 \).

Step by step solution

01

Identify the Relationship between Natural Logarithms and Logarithms Base 10

To express the natural logarithm \( \ln x \) in terms of base 10 logarithms \( \log_{10} x \), we can use the change of base formula for logarithms: \( \ln x = \log_{10} x \cdot \ln 10 \).
02

Apply the Change of Base Formula

Use the change of base formula to express \( \ln 3 \) as a base 10 logarithm: \( \ln 3 = \log_{10} 3 \cdot \ln 10 \). This becomes more useful when \( \ln 10 \) is considered a constant value.

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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 a crucial tool in expressing logarithms in different bases. Essentially, it allows you to convert any logarithm to another base, often making calculations simpler. Here's how it works:
  • For any positive number \(x\), the logarithm with base \(a\), \(\log_a x\), can be converted to a logarithm with a different base \(b\) using: \(\log_a x = \frac{\log_b x}{\log_b a}\).
In the context of natural logarithms (base \(e\)) and base 10 logarithms, the formula lets us express \(\ln x\) as \(\log_{10} x\) after multiplying by the constant \(\ln 10\). This transformation is useful because base 10 logarithms are more common in everyday applications and some computations. You often encounter them when working with calculators that default to base 10.
Logarithms Base 10
Logarithms base 10, often denoted as \(\log_{10}\), are called common logarithms. They are particularly widespread due to their simplicity and ease of use in computation.
  • The base 10 system aligns naturally with our decimal numbering system, making it very intuitive.
  • Base 10 logarithms help us easily simplify large numbers, as they scale down values efficiently.
  • Using common logarithms, many calculations in science, engineering, and other fields become more straightforward.
The change of base formula allows converting logarithms in other bases to base 10. This is handy because most calculators are designed to handle these common logarithms without additional steps. By expressing other logarithms via base 10, computations become more accessible and more familiar to work with.
Expressing Logarithms
Expressing logarithms involves writing logarithms of one base in terms of another. As seen earlier, using the change of base formula makes this possible.
  • This technique is especially needed when comparing values given in different logarithmic bases.
  • By expressing logarithms all in one base, such as base 10, we standardize and simplify our calculations.
When expressing \(\ln 3\) in terms of base 10 logarithms, the expression becomes \(\ln 3 = \log_{10} 3 \cdot \ln 10\). This transformation aligns the natural logarithm (base \(e\)) to a more familiar base for many applications, making results easier to interpret and consistent with commonly used logarithmic scales.

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Most popular questions from this chapter

Let \(\mathcal{N}=\mathcal{N}_{0 e^{k t}} .\) In this exercise we show that if \(\Delta t\) is very small, then \(\Delta \mathcal{N} / \Delta t \approx k \mathcal{N} .\) In other words, over very small intervals of time, the average rate of change of \(\mathcal{N}\) is proportional to \(\mathcal{N}\) itself. (a) Show that the average rate of change of the function \(\mathcal{N}=\mathcal{N}_{0} e^{t t}\) on the interval \([t, t+\Delta t]\) is given by $$\frac{\Delta \mathcal{N}}{\Delta t}=\frac{\mathcal{N}_{0} e^{k t}\left(e^{k \Delta t}-1\right)}{\Delta t}=\frac{\mathcal{N}\left(e^{k \Delta t}-1\right)}{\Delta t}$$ (b) In Exercise 26 of Section 5.2 we saw that \(e^{x} \approx x+1\) when \(x\) is close to zero. Thus, if \(\Delta t\) is sufficiently small, we have \(e^{k \Delta t} \approx k \Delta t+1 .\) Use this approximation and the result in part (a) to show that \(\Delta \mathcal{N} / \Delta t \approx k N\) when \(\Delta t\) is sufficiently close to zero.

Simplify each expression. (a) \(\ln e\) (b) \(\ln e^{-2}\) (c) \((\ln e)^{-2}\)

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$6\left(5-1.6^{x}\right) \geq 13$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\log _{2} \frac{2 x-1}{x-2}<0$$

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{3 x^{2}}=112$$
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