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Chapter 13: Problem 14

Use logarithms to the base 10 to find the indicated logarithms. $$\log _{100} 3720$$

Short Answer

Expert verified
Using the base 10 logarithm, \( \log_{100} 3720 \approx 1.7851 \).

Step by step solution

01

Change of Base Formula

To find a logarithm with a base other than 10, we can use the change of base formula. The change of base formula is given by:\[ \log_b a = \frac{\log_c a}{\log_c b} \]In this exercise, we are asked to find \( \log_{100} 3720 \), and we will use base 10 for our calculations. Thus, the expression becomes:\[ \log_{100} 3720 = \frac{\log_{10} 3720}{\log_{10} 100} \]
02

Calculate \(\log_{10} 100\)

Next, we need to calculate \( \log_{10} 100 \). We know that 100 can be written as \(10^2\). Therefore:\[ \log_{10} 100 = \log_{10} (10^2) = 2 \times \log_{10} 10 = 2 \times 1 = 2 \]
03

Calculate \( \log_{10} 3720 \)

Now, let's approximate \( \log_{10} 3720 \) using a scientific calculator. The result is approximately:\[ \log_{10} 3720 \approx 3.5702 \]
04

Apply the Change of Base Result

Now use the results from Steps 2 and 3 to find \( \log_{100} 3720 \) as follows:\[ \log_{100} 3720 = \frac{\log_{10} 3720}{\log_{10} 100} = \frac{3.5702}{2} \approx 1.7851 \]
05

Round the Answer

For final presentation, it's often helpful to round the answer to a certain number of decimal places to simplify communication. Therefore, rounding 1.7851 to 4 decimal places, we get approximately:\[ \log_{100} 3720 \approx 1.7851 \]

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Key Concepts

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

Change of Base Formula
When dealing with logarithms that don't have a base of 10, you can employ a neat trick called the Change of Base Formula. This formula is a lifesaver when your calculator only likes base 10 (common logarithms) or base e (natural logarithms). The formula goes like this:
  • Take the logarithm of your number using a preferred base, like 10.
  • Divide it by the logarithm of the original base, still using your preferred base.
In mathematical terms:\[\log_b a = \frac{\log_c a}{\log_c b}\]This expression helps you transform any log to a base that you can easily calculate, often base 10, enabling you to solve logarithmic expressions with ease. For instance, to compute \( \log_{100} 3720 \), use:\[\log_{100} 3720 = \frac{\log_{10} 3720}{\log_{10} 100}\]This makes the calculation far simpler, bringing everything into a manageable format.
Base 10 Logarithms
Base 10 logarithms, often referred to as common logarithms, are a favorite among mathematicians and students alike because they link directly to our decimal system. They are denoted by \( \log_{10} \) or sometimes just \( \log \).One of the neat properties of any base 10 logarithm is what happens with powers of 10:
  • \( \log_{10} 10 = 1 \)
  • \( \log_{10} 100 = 2 \)
  • \( \log_{10} 1000 = 3 \)
This shows that the logarithm tells you how many 10s you multiply together to get the number, which makes calculations like \( \log_{10} 100 = 2 \) straightforward. No wonder base 10 is the default log on calculators! These properties make it simple to approach logarithms in base 10, which is vital for further calculations like employing the Change of Base Formula.
Scientific Calculator Usage
A scientific calculator can be your best friend when dealing with logarithmic calculations, especially when you need to calculate logarithms to several decimal places. When using a scientific calculator:
  • Find the "log" button 鈥� it's for base 10 logs.
  • Enter the value you need the logarithm of, typically by typing the number first and then pressing the "log" button.
  • Read the result carefully; it may be a long decimal.
For instance, using a scientific calculator to find \( \log_{10} 3720 \), you'd perform these steps to approximate the value as 3.5702. Understanding how to correctly enter numbers and interpret results can help you seamlessly apply your findings to more advanced formulae like the Change of Base Formula.
Rounding Decimals
Rounding decimals is an essential skill for presenting your calculations in a clear and comprehensible manner. It helps to simplify long decimal numbers where excessive precision isn't necessary:
  • Identify how many decimal places are needed to express your result. Often, 4 places are sufficient for most school tasks.
  • Look at the number in the next decimal place to decide whether to round up or down 鈥� if it's 5 or more, round up.
  • Remember, rounding keeps your answer tidy and reader-friendly.
For example, when we calculated \( \log_{100} 3720 \), the result was approximately 1.7851. Since the fourth decimal place is important in standard rounding to four decimal places, this is the value to use in a neatly presented answer. Mastering rounding ensures your math results are both accurate and polished.

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Most popular questions from this chapter

Use a calculator to solve the given equations. $$\log 4 x+\log x=2$$

Show that the given functions are inverse functions of each other. Then display the graphs of each function and the line \(y=x\) on a graphing calculator and note that each is the mirror image of the other across \(y=x\). $$y=10^{x / 2} \text { and } y=2 \log _{10} x$$

$$\text {Plot the indicated graphs.}$$ One end of a very hot steel bar is sprayed with a stream of cool water. The rate of cooling \(R\) (in \(^{\circ} \mathrm{F} / \mathrm{s}\) ) as a function of the distance \(d\) (in in.) from one end of the bar is then measured, with the results shown in the following table. On log-log paper, plot \(R\) as a function of \(d .\) Such experiments are made to determine the hardness of steel. $$\begin{aligned} &\begin{array}{l|l|l|l|l} d \text { (in.) } & 0.063 & 0.13 & 0.19 & 0.25 \\ \hline R\left(^{\circ} \mathrm{F} / \mathrm{s}\right) & 600 & 190 & 100 & 72 \end{array}\\\ &\begin{array}{l|c|c|c|c|c} d \text { (in.) } & 0.38 & 0.50 & 0.75 & 1.0 & 1.5 \\ \hline R\left(^{\circ} \mathrm{F} / \mathrm{s}\right) & 46 & 29 & 17 & 10 & 6.0 \end{array} \end{aligned}$$

Use a calculator to verify the given values. $$2 \ln 6-\ln 3=\ln 12$$

Perform the indicated operations. The work \(W\) done by a sample of nitrogen gas during an isothermal (constant temperature) change from volume \(V_{1}\) to volume \(V_{2}\) is given by \(W=k \log _{e}\left(V_{2} / V_{1}\right) .\) Solve for \(V_{1}\).
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