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Solve for \(x .\) Hint: \(\log _{a} b=c \Leftrightarrow a^{c}=b\). $$ \log _{4} x=\frac{3}{2} $$

Short Answer

Expert verified
The solution is \( x = 8 \).

Step by step solution

01

Identify the Given Equation

We are given the equation \( \log_{4} x = \frac{3}{2} \). This expression describes a logarithmic relationship, meaning that 4 raised to a certain power will equal \( x \).
02

Apply the Definition of Logarithms

According to the hint, \( \log_{a} b = c \) implies \( a^{c} = b \). In our case, we have \( a = 4 \), \( b = x \), and \( c = \frac{3}{2} \). So we can write the equation as \( 4^{\frac{3}{2}} = x \).
03

Calculate \(4^{\frac{3}{2}}\)

The expression \(4^{\frac{3}{2}}\) can be simplified by rewriting it as \( (4^{1})^{\frac{3}{2}} = \left( 4^{\frac{1}{2}} \right)^{3} \). Since \( 4^{\frac{1}{2}} = \sqrt{4} = 2 \), so \( 2^{3} = 8 \).
04

Conclude the Solution

From our calculations, we find that \( x = 8 \). Therefore, the value of \( x \) that satisfies the original equation \( \log_{4} x = \frac{3}{2} \) is 8.

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Key Concepts

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

Properties of Logarithms
Understanding logarithms is crucial for solving equations like the one given in the problem. A logarithm answers the question: "To what exponent must we raise a given base to obtain a certain number?" The general form of a logarithmic expression is \( \log_{a} b = c \), where \( a \) is the base, \( b \) is the result, and \( c \) is the exponent. This can be expressed in exponential form, as \( a^{c} = b \). This property is pivotal when converting logarithms to exponential equations, allowing for easier computation of unknowns.

Some important properties of logarithms include:
  • Product property: \( \log_{a}(mn) = \log_{a} m + \log_{a} n \)
  • Quotient property: \( \log_{a}\left(\frac{m}{n}\right) = \log_{a} m - \log_{a} n \)
  • Power property: \( \log_{a}(m^n) = n \log_{a} m \)

These properties help simplify complex logarithmic expressions and are handy tools for manipulating and solving equations.
Exponential Functions
Exponential functions often appear in the context of solving logarithmic equations. An exponential function is any function of the form \( a^x \), where \( a \) is a constant and \( x \) is a variable. They are characterized by their rapid growth or decay, depending on whether the base is greater than or less than one.

In the context of the original problem \( \log_{4} x = \frac{3}{2} \), we can transform the equation into an exponential form, as explained in the solution. This involves recognizing that the logarithm can "invert" to find the power needed to achieve \( x \).

By applying the definition, we turn the logarithmic equation into \( 4^{\frac{3}{2}} = x \). The computation then becomes focused on evaluating this exponential expression. This step is critical, showing how logarithms and exponentials are two sides of the same coin, allowing us to shift between forms to solve equations.
Solving Equations
The process of solving logarithmic equations like \( \log_{4} x = \frac{3}{2} \) involves several steps that require a good grasp of logarithmic properties and exponential functions.

To begin, you identify the logarithmic form of the equation and then convert it into an exponential form, as shown in the example from the original exercise. The conversion is guided by the definition \( \log_{a} b = c \), which implies \( a^{c} = b \).

After transforming the original equation into the form \( 4^{\frac{3}{2}} = x \), the next step is to solve for \( x \), which involves calculating the exponential expression. This particular calculation, \( 4^{\frac{3}{2}} \), can be simplified to \( (4^{\frac{1}{2}})^3 \). First, you determine \( 4^{\frac{1}{2}} = 2 \) because \( 4^{\frac{1}{2}} \) is the square root of 4. Then, \( 2^3 = 8 \), leading to the solution \( x = 8 \).

These steps demonstrate that by using properties of logarithms and understanding exponential functions, we can make complex equations more manageable and find solutions efficiently.

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