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Use the various properties of exponential and logarithmic functions to evaluate the expressions in parts \((a)-(c)\) $$\text { Given } f(x)=3^{x}, \text { find } \quad \text { (a) } f\left(\log _{3} 2\right) \text { (b) } f\left(\log _{3}(\ln 3)\right) \text { (c) } f\left(\log _{3}(2 \ln 3)\right)$$

Short Answer

Expert verified
(a) 2, (b) \( \ln 3 \), (c) \( 2 \ln 3 \)

Step by step solution

01

Understand the Function

Given the function is defined as \( f(x) = 3^x \). We need to evaluate this function for different inputs provided in parts (a), (b), and (c).
02

Evaluate Part (a)

For part (a), we need to find \( f(\log_{3} 2) \). Substituting \( \log_{3} 2 \) into the function, we get: \[ f(\log_{3} 2) = 3^{\log_{3} 2} \]. Using the property of logarithms that \( a^{\log_{a} b} = b \), we find that \( 3^{\log_{3} 2} = 2 \). So, \( f(\log_{3} 2) = 2 \).
03

Evaluate Part (b)

For part (b), we need to find \( f(\log_{3}(\ln 3)) \). Substituting \( \log_{3}(\ln 3) \) into the function, we get: \[ f(\log_{3}(\ln 3)) = 3^{\log_{3}(\ln 3)} \]. Using the property of logarithms, \( 3^{\log_{3}(\ln 3)} = \ln 3 \). So, \( f(\log_{3}(\ln 3)) = \ln 3 \).
04

Evaluate Part (c)

For part (c), we need to find \( f(\log_{3}(2 \ln 3)) \). Substituting \( \log_{3}(2 \ln 3) \) into the function, we get: \[ f(\log_{3}(2 \ln 3)) = 3^{\log_{3}(2 \ln 3)} \]. Using the property of logarithms, \( 3^{\log_{3}(2 \ln 3)} = 2 \ln 3 \). So, \( f(\log_{3}(2 \ln 3)) = 2 \ln 3 \).

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Key Concepts

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

Properties of Logarithms
Logarithms are the inverse functions of exponentials. To work with logarithms effectively, it's crucial to understand some of their key properties:

1. **Basic Property**: If you have \(a^{\log_{a} b} = b\), it simplifies logarithmic expressions involving exponentiation.
2. **Product Rule**:\(\log_{a}(xy) = \log_{a}(x) + \log_{a}(y)\)
3. **Quotient Rule**: \(\log_{a}\left(\frac{x}{y}\right) = \log_{a}(x) - \log_{a}(y)\)
4. **Power Rule**:\(\log_{a}(x^k) = k \cdot \log_{a}(x)\)
5. **Change of Base Rule**: \(\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}\)

In our original exercise, we frequently use the basic property that \(a^{\log_{a}(x)} = x\). For example, in step (a) of the solution, \( 3^{\log_{3} 2} = 2\), leveraging the fact that the base raised to its own logarithm of any number returns that number.
Understanding these properties simplifies complex expressions and is essential for solving logarithmic and exponential problems.
Exponential Functions
Exponential functions involve a constant base being raised to a variable exponent. For example, in the function \(f(x) = 3^x\), the base is 3.

1. **Growth Rate**: Exponential functions grow rapidly because the rate of growth is proportional to the function’s current value.
2. **Inverse Relationship**: The inverse of an exponential function is a logarithmic function, hence \(y = 3^x\) is the inverse of \(y = \log_{3}(x)\).
3. **Properties**: They follow specific rules. For example, \(a^{m+n} = a^m \cdot a^n\) and \((a^m)^n = a^{mn}\). These properties help simplify and understand how exponential expressions can be manipulated.

In the original problem, we evaluated \(f(x)\) by substituting logarithmic expressions into the exponential function. In part (a), substituting \(\log_{3} 2\) into \(3^x\) simplifies directly to 2 due to the \(a^{\log_{a} b} = b\) property.
Logarithmic Evaluations
Evaluating logarithmic expressions often involves simplifying the expressions using known logarithmic properties.

1. **Direct Evaluations**: Sometimes the expressions can be directly calculated using properties of logarithms. For example, \(\log_{3} 3 = 1\) because \(3^1 = 3\).
2. **Combining Logarithms**: You can either combine or break down logarithmic expressions using product, quotient, and power rules. For example, \(\log_{3}(2 \ln 3)\) can be simplified using the product rule and constant multipliers.
3. **Changing the Base**: This rule is useful if the logarithm’s base is different from the required. You can change the base to make the calculation easier.

In the solutions provided, logarithmic evaluations simplified the process of finding \(f(x)\). For instance, in part (c) \( f(\log_{3}(2 \ln 3)) = 3^{\log_{3}(2 \ln 3)} = 2 \ln 3 \) transformed the expression into an easily understandable form and gave the final result straightforwardly.

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Most popular questions from this chapter

World Population Growth since 2000 , world population in millions closely fits the exponential function $$ y=6084 e^{00120 x} $$ where \(x\) is the number of years since 2000 . (Source: U.S. Census Bureau.) (a) The world population was about 6853 million in 2010 . How closely does the function approximate this value? (b) Use this model to predict the population in 2020 . (c) Use this model to predict the population in 2030 . (d) Explain why this model may not be accurate for 2030 .

Use the change-of-base theorem to find an approximation to four decimal places for each logarithm. $$\log _{1 / 3} 2$$

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Use another type of logistic function. Tree Growth The height of a certain tree in feet after \(x\) years is modeled by $$ f(x)=\frac{50}{1+47.5 e^{-0.22 x}} $$ (a) Make a table for \(f\) starting at \(x=10,\) and incrementing by \(10 .\) What appears to be the maximum height of the tree? (b) Graph \(f\) and identify the horizontal asymptote. Explain its significance. (c) After how long was the tree 30 ft tall?

Decide whether the given functions are inverses. $$f=\\{(1,1),(3,3),(5,5)\\} ; \quad g=\\{(1,1),(3,3),(5,5)\\}$$

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