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Complete the proof of the logarithmic property \(\log _{a} u v=\log _{a} u+\log _{a} v\) Let \(\log _{a} u=x\) and \(\log _{a} v=y\). \(a^{x}=\quad\) and \(a^{y}=\quad\) Rewrite in exponential form. \(u \cdot v=\quad \cdot \quad=a \quad\) Multiply and substitute for \(u\) and \(v\). \(=x+y\) Rewrite in logarithmic form. \(\log _{a} u v=\quad+\) Substitute for \(x\) and \(y\).

Short Answer

Expert verified
The proof of the logarithmic property \(\log _{a} u v=\log _{a} u+\log _{a} v\) has been successfully completed by using the basic properties of exponentials and logarithms. By converting logarithms into an exponential form, applying the rules of exponentiation, rewriting the equation in logarithmic form, and substituting the denoted variables, we can prove the desired property.

Step by step solution

01

Converting Logarithmic to Exponential form

Let's start by converting the given logarithmic forms to exponential forms. Given \(\log _{a} u=x\) and \(\log _{a} v=y\), these can be converted to \(a^{x}=u\) and \(a^{y}=v\) respectively.
02

Applying the Exponentiation Rule

Using the rule of exponentiation which states \(a^{x+y}=a^x \cdot a^y\), we can express \(a^{x + y}\) as \(u \cdot v\) since \(a^x\) is equal to \(u\) and \(a^y\) is equal to \(v\). Thus, \(u \cdot v = a^{x + y}\).
03

Rewriting in Logarithmic Form

Having established that \(u \cdot v = a^{x + y}\), one can transcribe this back to the logarithmic form as \(\log_a {u \cdot v} = x + y\).
04

Substituting for x and y

We can now replace \(x\) and \(y\) with \(\log _{a} u\) and \(\log _{a} v\) respectively in \(\log _{a} {u \cdot v} = x + y\) which gives us our desired result \(\log _{a} u v=\log _{a} u+\log _{a} v\).

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Key Concepts

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

exponential form
In mathematics, exponential form is a way to express repeated multiplication of a number by itself. It is often used in situations where a number is raised to a particular power. The base of the exponential expression is the number being multiplied, and the exponent indicates how many times the base is to be multiplied by itself.

For example, given the expressions \(\log_a u = x\) and \(\log_a v = y\), these can be rewritten in exponential form as \(a^x = u\) and \(a^y = v\). This is because the definition of a logarithm implies that the logarithm is the exponent to which the base must be raised to yield a particular number. This rewrite from logarithmic to exponential form helps in understanding problems that involve multiplication or division of powers.
  • The base \(a\) raised to the power of \(x\) gives us \(u\).
  • The base \(a\) raised to the power of \(y\) gives us \(v\).
This conversion is crucial as it sets up further manipulations like multiplying or factoring exponentiated terms, which often appear in more complex mathematical calculations.
logarithmic form
Logarithmic form is an expression of exponential equations which shows the power to which a base must be raised to obtain a certain number. It helps in solving equations where the unknown is an exponent.

Given an expression like \(u \cdot v = a^{x + y}\), we can express this in logarithmic form as \(\log_a (u \cdot v) = x + y\). This step uses the concept that if two quantities are equal, their logarithms (with the same base) are also equal.
  • This form is particularly useful for simplifying expressions where the product of numbers is involved.
  • Having both products and sums in logarithmic expressions allows the application of various logarithmic properties for further simplification.
Rewriting expressions in logarithmic form is a powerful algebraic technique, as it often makes it easier to apply logarithmic properties and solve more complicated problems efficiently.
logarithm multiplication property
The logarithm multiplication property is a fundamental rule in logarithms, which states that the logarithm of a product is equal to the sum of the logarithms of the factors.

Mathematically, it is expressed as:\[ \log_a (u \, v) = \log_a u + \log_a v \]This property can be understood by considering the relationships established in converting between exponential and logarithmic forms. From \(u \cdot v = a^{x+y}\), since \(x = \log_a u\) and \(y = \log_a v\), substituting back gives us the multiplication property of logarithms.
  • It's very useful in simplifying logarithmic expressions, breaking down complex products into manageable parts.
  • This property aids in various fields like data analysis or solving polynomial equations, where products need to be simplified into sums for easier computation.
Understanding this property helps decompose complex numbers into simpler terms, making problem-solving much more accessible for everyone from students to professional mathematicians.

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Most popular questions from this chapter

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 \(2006 ?\)

In a group project on learning theory, a mathematical model for the percent \(P\) (in decimal form) of correct responses after \(n\) trials was found to be \(P=\frac{0.98}{1+e^{-0.3 n}}, \quad n \geq 0\) (a) After how many trials will \(80 \%\) of the responses be correct? (That is, for what value of \(n\) will \(P=0.8\) ?) (b) Use a graphing utility to graph the memory model and confirm the result found in part (a). (c) Write a paragraph describing the memory model.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(8\left(4^{6-2 x}\right)+13=41\)

Use a graphing utility to solve the equation. Approximate the result to three decimal places. Verify your result algebraically.\(2^{x}-7=0\)

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

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