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Chapter 4: Problem 23

Rewrite the expression in terms of \(\ln 4\) and \(\ln 5 .\),$$\ln \frac{25}{4}$$.

Short Answer

Expert verified
The expression \( \ln \frac{25}{4} \) rewritten in terms of \( \ln 4 \) and \( \ln 5 \) is \( 2 \ln 5 - \ln 4 \).

Step by step solution

01

Rewrite Division as Subtraction

Apply the division rule for logarithms, which states that the logarithm of a quotient is the difference of the logarithms i.e., \( \ln(a/b) = \ln a - \ln b \). Therefore, \( \ln \frac{25}{4} \) is rewritten as \( \ln 25 - \ln 4 \).
02

Apply Exponent Rule to \( \ln 25 \)

We know that 25 equals 5 squared, further more, rule of logarithms states that the logarithm of an exponent is equal to that exponent times logarithm of the base. Hence, \( \ln 25 \) can be rewritten as \( 2 \ln 5 \), this will allow us to express our expression using only \( \ln 4 \) and \( \ln 5 \).
03

Insert results

Now, insert the obtained results from step 1 and step 2 back into the expression. Replace \( \ln 25 - \ln 4 \) in step 1 with \( 2 \ln 5 - \ln 4 \). Thus, \( \ln \frac{25}{4} \) is equal to \( 2 \ln 5 - \ln 4 \).

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Key Concepts

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

Logarithmic Properties
Logarithmic properties are fundamental concepts that help simplify and manipulate logarithmic expressions. They are similar to laws of algebra but tailored for logarithms. Understanding these properties allows us to transform log expressions, solve log equations, and make computations more manageable.
  • Product Property: This states that the logarithm of a product is the sum of the logarithms. For example, \( \ln (a \cdot b) = \ln a + \ln b \). This property helps in expanding logarithms when you're multiplying values within the log.
  • Quotient Property: It asserts that the logarithm of a quotient is the difference of the logarithms, such as \( \ln \frac{a}{b} = \ln a - \ln b \). This particular property was used in the original problem to break down the expression \( \ln \frac{25}{4} \).
  • Power Property: It guides us on how to deal with exponents inside a logarithm. The power property shows that the logarithm of an expression with an exponent can be written as the exponent multiplied by the logarithm of the base, that is \( \ln (a^b) = b \cdot \ln a \).
  • Change of Base Formula: While not used in the current solution, it allows us to convert logs from one base to another, which is \( \log_b a = \frac{\log_k a}{\log_k b} \) for any positive base \( k \).

These properties are essential for understanding how logarithms simplify complex mathematical expressions and solve equations.
Exponent Rule
The exponent rule for logarithms is a crucial property utilized to deal with powers within logarithmic expressions. This rule states that when you have a logarithm of a number raised to an exponent, you can bring the exponent down in front of the logarithm.

For example, utilizing the exponent rule, \( \ln (a^b) \) is simplified to \( b \cdot \ln a \). This transformation happens because raising a number to a power inside a logarithm is essentially multiple multiplications, and logarithms convert those into addition, simplifying the computation.
  • In the original problem, this rule was applied to rewrite \( \ln 25 \). Recognizing that 25 is the same as \( 5^2 \), the exponent rule simplifies \( \ln 25 \) to \( 2 \cdot \ln 5 \).
  • This helps reframe the problem so all terms are in the simplest terms of \( \ln 5 \) and \( \ln 4 \).

The exponent rule is essential for manipulating expressions within logarithms to make them more accessible and easier to work with.
Division Rule for Logarithms
The division rule for logarithms is another vital property that helps in simplifying logarithmic expressions, specifically those containing division. As it's defined, the logarithm of a quotient is the difference between the logarithms of the numerator and the denominator.

This rule is represented as \( \ln \frac{a}{b} = \ln a - \ln b \), allowing us to split a complex log expression into simpler parts by subtracting one log from another.
  • This property is helpful when dealing with expressions that need to be simplified or expressed in terms of other known logarithms.
  • In the provided problem solution, the division rule was used to break down \( \ln \frac{25}{4} \) into \( \ln 25 - \ln 4 \), which then allowed the use of the exponent rule to express it in terms of \( \ln 4 \) and \( \ln 5 \).
  • By using this property, you can often transform logarithmic expressions into a form that is much easier to read and solve, providing a clearer path to the solution.

Mastering the division rule for logarithms can significantly enhance your skills in solving algebraic expressions that involve logarithms.

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Most popular questions from this chapter

Sketch the graph of the function. $$f(x)=|x-2|-8$$

A beaker of liquid at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C}\) The temperature of the liquid is measured every 5 minutes for a period of \(\frac{1}{2}\) hour. The results are recorded in the table, where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$\begin{array}{|c|c|}\hline \text { Time, } t 0 & \text { Temperature, } T\\\\\hline 0 & 78.0^{\circ} \\\5 & 66.0^{\circ} \\\10 & 57.5^{\circ} \\\15 & 51.2^{\circ} \\\20 & 46.3^{\circ} \\\25 & 42.5^{\circ} \\\30 & 39.6^{\circ} \\\\\hline\end{array}$$ (a) Use the regression feature of a graphing utility to find a linear model for the data. Use the graphing utility to plot the data and graph the model in the same viewing window. Do the data appear linear? Explain. (b) Use the regression feature of the graphing utility to find a quadratic model for the data. Use the graphing utility to plot the data and graph the model in the same viewing window. Do the data appear quadratic? Even though the quadratic model appears to be a good fit, explain why it might not be a good model for predicting the temperature of the liquid when \(t=60.\) (c) The graph of the temperature of the room should be an asymptote of the graph of the model. Subtract the room temperature from each of the temperatures in the table. Use the regression feature of the graphing utility to find an exponential model for the revised data. Add the room temperature to this model. Use the graphing utility to plot the original data and graph the model in the same viewing window. (d) Explain why the procedure in part (c) was necessary for finding the exponential model.

The yield \(V\) (in millions of cubic feet per acre) for a forest at age \(t\) years is given by \(V=6.7 e^{-48.1 / t}\) (a) Use a graphing utility to graph the function. (b) Determine the horizontal asymptote of the function. Interpret its meaning in the context of the problem. (c) Find the time necessary to obtain a yield of 1.3 million cubic feet.

(a) complete the table to find an interval containing the solution of the equation, (b) use a graphing utility to graph both sides of the equation to estimate the solution, and (c) solve the equation algebraically. Round your results to three decimal places. $$5 \log _{10}(x-2)=11$$ $$\begin{array}{|l|l|l|l|l|l|}\hline x & 150 & 155 & 160 & 165 & 170 \\\\\hline 5 \log _{10}(x-2) & & & & & \\\\\hline\end{array}$$

Evaluate the function for \(f(x)=3 x+2\) and \(g(x)=x^{3}-1.\) $$(f-g)(-1)$$
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