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If \(f(x)=a^{x}\) and \(f(3)=27,\) find each value. (a) \(f(1)\) (b) \(f(-1)\) (c) \(f(2)\) (d) \(f(0)\)

Short Answer

Expert verified
(a) 3, (b) \(\frac{1}{3}\), (c) 9, (d) 1

Step by step solution

01

Identify Known Information

We are given the function \(f(x) = a^x\) and the value \(f(3) = 27\). This means that when \(x = 3\), \(a^3 = 27\).
02

Solve for the Base 'a'

To find \(a\), solve the equation \(a^3 = 27\). Since \(27 = 3^3\), it follows that \(a = 3\).
03

Calculate f(1)

Now that \(a=3\), substitute \(x=1\) into the function: \(f(1) = 3^1 = 3\).
04

Calculate f(-1)

Substitute \(x=-1\) into the function: \(f(-1) = 3^{-1} = \frac{1}{3}\).
05

Calculate f(2)

Substitute \(x=2\) into the function: \(f(2) = 3^2 = 9\).
06

Calculate f(0)

Substitute \(x=0\) into the function: \(f(0) = 3^0 = 1\).

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Key Concepts

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

Evaluating Functions
Evaluating functions is an essential skill in mathematics that involves finding the value of a function for a given input. In this case, we have the function \( f(x) = a^x \). Evaluating a function means that we are tasked with determining the output \( f(x) \) when we plug a specific value of \( x \) into the expression for the function.

For instance, consider the example where we're asked to find \( f(1) \). Once we know the base \( a \), which in our example turns out to be 3, we can evaluate it by simply calculating \( f(1) = 3^1 = 3 \).

When evaluating functions, make sure you substitute the input value correctly in all occurrences of the variable. This step involves both understanding the function's formula and following the correct order of operations. Each step builds your understanding, and through practice, you'll become more comfortable with evaluating different kinds of functions.
Power and Exponent Rules
Power and exponent rules are foundational principles in algebra that dictate how to handle powers of numbers and expressions. Exponents signify repeated multiplication of a base number, and understanding how to manipulate them is crucial.

Few key rules include:
  • Power of a Power: \((a^m)^n = a^{m \times n}\)
  • Multiplying Same Bases: \(a^m \times a^n = a^{m+n}\)
  • Dividing Same Bases: \(a^m \div a^n = a^{m-n}\)
  • Negative Exponent: \(a^{-n} = \frac{1}{a^n}\)
  • Zero Exponent: \(a^0 = 1\) (where \(a eq 0\))
Understanding these rules makes calculating powers like \(3^1\), \(3^{-1}\), or \(3^2\) more intuitive. For the problem given:
- \(3^1\) uses the identity property, resulting in 3.
- \(3^{-1}\) demonstrates the negative exponent rule, which becomes \(\frac{1}{3}\).
- \(3^0\) uses the zero exponent rule, giving us a result of 1.
These rules are vital not only in academic exercises but also in advanced computations, as they help simplify expressions and solve equations more efficiently.
Solving Equations
Solving equations involves finding the value of the variable that makes the equation true. In the context of exponential equations like \(a^3 = 27\), solving such an equation requires a keen understanding of powers and roots.

Here, we need to find the base \(a\) when raised to the power of 3 equals 27. Recognizing that \(27 = 3^3\), we discern that \(a = 3\).

To solve such equations:
  • Express one side as a power if possible. This often involves recognizing potent numbers (like 8, 27, 64) as they can be expressed as powers of smaller integers.
  • Compare the exponents and solve for the base if the bases and their exponents on both sides are equal.
  • Use roots if an equation is not immediately clear.
In essence, solving equations, especially involving exponentials, combines several skills: breaking down numbers into their power components and applying algebraic manipulation to isolate the variable. By practicing these steps, problem-solving becomes logical and more approachable.

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Most popular questions from this chapter

Use the properties of logarithms to rewrite each expression as a single logarithm with coefficient 1 . Assume that all variables represent positive real numbers. $$-\frac{2}{3} \log _{5} 5 m^{2}+\frac{1}{2} \log _{5} 25 m^{2}$$

For each function that is one-to-one, write an equation for the inverse function of \(y=f(x)\) in the form \(y=f^{-1}(x),\) and then graph \(f\) and \(f^{-1}\) on the same axes. Give the domain and range of \(f\) and \(f^{-1} .\) If the function is not one-to-one, say so. $$y=\frac{4}{x}$$

The given function \(f\) is one-to-one. Find \(f^{-1}(x)\). $$f(x)=\sqrt{x^{2}-4}, x \geq 2$$

Although a function may not be one-to-one when defined over its "natural" domain, it may be possible to restrict the domain in such a way that it is one-to-one and the range of the function is unchanged. For example, if we nestrict the domain of the function \(f(x)=x^{2}\) (which is not one-to-one over \((-\infty, \infty)\) to \([0, \infty)\), we obtain a one-to-one function whose range is still \([0, \infty)\) See the figure to the right. Notice that we could choose to restrict the domain of \(f(x)=x^{2}\) to \((-\infty, 0]\) and also obtain the graph of a one-to-one function, except that it would be the left half of the parabola. For each function in Exercises \(117-122\), restrict the domain so that the function is one-to-one and the range is not changed. You may wish to use a graph to help decide. Answers may vary. (GRAPHS CANNOT COPY) $$f(x)=-\sqrt{x^{2}-16}$$

Answer each of the following. If \(f\) is a function that has an inverse and the graph of \(f\) lies completely within the second quadrant, then the graph of \(f^{-1}\) lies completely within the __________ quadrant.

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