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Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} \sqrt{3}\)

Short Answer

Expert verified
The result of the logarithmic function is approximately \(0.248\).

Step by step solution

01

Rewrite the square root as an exponent

Rewrite \(\sqrt{3}\) as \(3^{0.5}\). So the expression they need to evaluate becomes \(\log _{10} 3^{0.5}\)
02

Use the property of logarithms

When you apply the property of logarithms \(a \log b = \log b^a\), the result is \(0.5 \log _{10} 3\).
03

Enter into the calculator

Input the expression \(0.5 \log _{10} 3\) into your calculator and round your result to three decimal places. The answer will be the decimal you obtain.

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Key Concepts

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

Logarithms
Understanding logarithms is essential for solving many mathematical problems involving exponential relationships. A logarithm answers the question: To what power must a given base number be raised to obtain a specific number? The expression \( \log_b x \) corresponds to the power or exponent \( y \) in the equation \( b^y = x \). In the exercise \( \log_{10} \sqrt{3} \) we're looking for the power to which 10 must be raised to get \( \sqrt{3} \).

Logarithms come in handy when dealing with large numbers or complex calculations since they transform multiplicative processes into additive ones. This transformation is possible due to the unique properties of logarithms, which will be discussed in a following section. They simplify the calculations and make it easier to understand the underlying relationships. Regular practice with a logarithm calculator will reinforce your comprehension and application of this concept.
Exponential Expressions
Exponential expressions are a way to represent repeated multiplication. The expression \( a^b \) signifies that the base \( a \) is multiplied by itself \( b \) times. In the context of logarithms, understanding exponential expressions is crucial because logarithms are, in essence, the inverse of exponentials.

In the given problem, the square root of 3, \( \sqrt{3} \) can be expressed as \( 3^{0.5} \) because square roots correspond to the one-half power. When you use a logarithm calculator, rewriting square roots or other radical expressions as exponents simplifies the computation. It's crucial to grasp this idea because it consistently appears in problems involving growth and decay, compounding interest, and many areas of science and engineering.
Properties of Logarithms
The properties of logarithms greatly facilitate the process of calculation and allow us to manipulate logarithmic expressions. One of these properties is the power rule, which states that \( \log_b (a^c) = c \cdot \log_b a \). It allows us to move the exponent of the argument to the front of the logarithm, effectively converting a multiplicative relationship into an additive one.

This very property is applied in Step 2 of the textbook solution, where the expression \( \log_{10} 3^{0.5} \) becomes \( 0.5 \cdot \log_{10} 3 \). Utilizing these properties is fundamental for simplifying logarithmic expressions before inputting them into calculators. Recognizing and applying the properties correctly can transform a seemingly complex calculation into a more manageable form, making it easier to input and reducing potential errors during calculation.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(3 \ln 5 x=10\)

The value \(y\) (in billions of dollars) of U.S. currency in circulation (outside the U.S. Treasury and not held by banks) from 1996 to 2005 can be approximated by the model \(y=-302+374 \ln t, \quad 6 \leq t \leq 15\) where \(t\) represents the year, with \(t=6\) corresponding to 1996\. (Source: Board of Governors of the Federal Reserve System) (a) Use a graphing utility to graph the model. (b) Use a graphing utility to estimate the year when the value of U.S. currency in circulation exceeded \(\$ 600\) billion. (c) Verify your answer to part (b) algebraically.

Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0875\)

Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0725\)

Population The population \(P\) of the United States officially reached 300 million at about 7:46 A.M. E.S.T. on Tuesday, October 17,2006 . The table shows the U.S. populations (in millions) since 1900. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1900 & 76 \\ \hline 1910 & 92 \\ \hline 1920 & 106 \\ \hline 1930 & 123 \\ \hline 1940 & 132 \\ \hline 1950 & 151 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1960 & 179 \\ \hline 1970 & 203 \\ \hline 1980 & 227 \\ \hline 1990 & 250 \\ \hline 2000 & 282 \\ \hline 2006 & 300 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1900 . (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Graph the exponential model in base \(e\) with the scatter plot of the data. What appears to be happening to the relationship between the data points and the regression curve at \(t=100\) and \(t=106 ?\) (d) Use the regression feature of a graphing utility to find a logistic growth model for the data. Graph each model using the window settings shown below. Which model do you think will give more accurate predictions of the population well beyond \(2006 ?\)

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