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Chapter 6: Problem 24

Solve each equation. Give the exact answer. $$\log _{4} x=-\frac{1}{6}$$

Short Answer

Expert verified
\( x = 2^{-1/3} \)

Step by step solution

01

Understand the Logarithm Equation

The given equation is \( \log_{4} x = -\frac{1}{6} \). This means that we are searching for a number \( x \) such that the base 4 logarithm of \( x \) equals \(-\frac{1}{6}\).
02

Convert Logarithm to Exponential Form

To convert the logarithmic equation into an exponential equation, use the definition of logarithms: for \( \log_{b} a = c \), it is equivalent to \( b^{c} = a \). For our equation, \( 4^{-\frac{1}{6}} = x \).
03

Simplify the Expression

Calculate \( 4^{-\frac{1}{6}} \). This is the same as \( \frac{1}{4^{\frac{1}{6}}} \). Since \( 4 = 2^{2} \), it further simplifies to \( 2^{-\frac{1}{3}} \).
04

Express the Answer

Thus, the expression for \( x \) is \( x = 2^{-1/3} \). This is the exact answer to the equation given.

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Key Concepts

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

Exponential Form
When we talk about exponential form, we are essentially converting an equation from one mathematical expression to another equivalent form. It's common in mathematics where we deal with powers and roots. In the context of logarithmic equations, converting to exponential form is a crucial step.
To convert a logarithmic expression into an exponential one:
  • Start with the given equation \( \log_{b} a = c \).
  • According to the definition of logarithms, this can be rewritten as \( b^{c} = a \).
This means if we know the base (\(b\)), exponent (\(c\)), and the result (\(a\)), we can switch between these forms.
In our example, \( \log_{4} x = -\frac{1}{6} \) becomes \( 4^{-\frac{1}{6}} = x \) once converted to exponential form. This transformation helps in simplifications and calculations.
Properties of Exponents
Exponents have some important properties that allow us to simplify expressions effectively. Understanding these can help greatly in solving equations and in manipulating algebraic expressions.
Some of the key properties of exponents include:
  • \( a^{m} \times a^{n} = a^{m+n} \) - When multiplying like bases, add the exponents.
  • \( \frac{a^{m}}{a^{n}} = a^{m-n} \) - When dividing like bases, subtract the exponents.
  • \( (a^{m})^{n} = a^{m\cdot n} \) - When raising an exponent to another power, multiply the exponents.
  • \( a^{-m} = \frac{1}{a^{m}} \) - A negative exponent means reciprocal.
  • \( a^{0} = 1 \) - Any non-zero base raised to the zero power is 1.
In the solution provided, converting \( 4^{-\frac{1}{6}} \) involves using the negative exponent property: \( 4^{-\frac{1}{6}} = \frac{1}{4^{\frac{1}{6}}} \). Then, recognizing that \( 4 = 2^{2} \), we can apply further exponent rules to simplify.
Definition of Logarithms
Logarithms are the inverse operation of exponentiation. This means they help us find which power a number (called the base) must be raised to, to arrive at another number.
For example, if \( a^{b} = c \), then \( \log_{a} c = b \). This definition is pivotal for rewriting logarithmic equations into exponential form, aiding in problem-solving.
Logarithms have some unique properties:
  • \( \log_{a}(xy) = \log_{a}(x) + \log_{a}(y) \) - The logarithm of a product is the sum of the logarithms.
  • \( \log_{a}\left(\frac{x}{y}\right) = \log_{a}(x) - \log_{a}(y) \) - The logarithm of a quotient is the difference.
  • \( \log_{a}(x^{n}) = n\log_{a}(x) \) - The logarithm of a power is the exponent times the logarithm.
  • \( \log_{a}(1) = 0 \) - The logarithm of 1 is always 0.
  • \( \log_{a}(a) = 1 \) - The logarithm of the base is always 1.
In the original problem, \( \log_{4} x = -\frac{1}{6} \), using the definition of a logarithm allows us to express it as \( 4^{-\frac{1}{6}} = x \), simplifying our path to the solution.

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Most popular questions from this chapter

In general, it is not possible to find exact solutions analytically for equations that involve exponential or logarithmic functions together with polynomial, radical, and rational functions. Solve each equation using a graphical method, and express solutions to the nearest thousandth if an approximation is appropriate. $$3^{-x}=\sqrt{x+5}$$

Use the definition of inverse functions to show analytically that \(f\) and \(g\) are inverses. $$f(x)=-x^{7}, \quad g(x)=-\sqrt[7]{x}$$

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) $$f(x)=-\sqrt{x^{2}-16}$$

Use a graphing calculator to solve each equation. Give solutions to the nearest hundredth. $$e^{3 x}=x^{3}+4$$

Use the definition of inverse functions to show analytically that \(f\) and \(g\) are inverses. $$f(x)=4 x+3, \quad g(x)=\frac{x-3}{4}$$
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