Problem 8
Question
$$ \log _{4} 4^{x} $$
Step-by-Step Solution
Verified Answer
The expression simplifies to \( x \).
1Step 1: Recall the Logarithm Power Rule
The logarithm power rule states that for any positive number \( b \), exponent \( x \), and logarithm base \( a \), the expression \( \log_a(b^x) \) simplifies to \( x \cdot \log_a(b) \). In symbols, \( \log_a(b^x) = x \cdot \log_a(b) \).
2Step 2: Apply the Logarithm Power Rule
Apply the logarithm power rule to the given exercise \( \log_{4}(4^x) \). According to the rule, it simplifies to:\[ x \cdot \log_{4}(4) \]
3Step 3: Recognize Simplification of Log Base
The property of logarithms \( \log_{a}(a) = 1 \) applies here because the base of the logarithm and the base of the exponent are the same. Therefore, \( \log_4(4) = 1 \).
4Step 4: Final Simplification
Substitute \( \log_4(4) = 1 \) into the expression obtained:\[ x \cdot 1 = x \]
5Step 5: Conclusion
Therefore, the expression \( \log_{4}(4^x) \) simplifies to \( x \).
Key Concepts
Logarithm Power RuleLogarithmic SimplificationLogarithm Base Property
Logarithm Power Rule
When dealing with logarithms, one useful property is the Logarithm Power Rule. This rule makes it easier to work with expressions where numbers are raised to a power. The power rule states: if you have an expression of the form \( \log_a(b^x) \), it simplifies to \( x \cdot \log_a(b) \).
Here's why this works: Logarithms are the inverse operations of exponentiation. When you use a logarithm, you're essentially asking: "to what power must I raise the base \( a \) to get \( b^x \)?". If \( b \) is raised to \( x \), you're essentially counting \( x \) times, leading to the straight multiplication by \( x \).
This rule is particularly handy because it allows us to "pull down" the exponent and convert a more complex exponential expression into a simpler product. This simplification can make subsequent calculations much easier and faster to execute.
Here's why this works: Logarithms are the inverse operations of exponentiation. When you use a logarithm, you're essentially asking: "to what power must I raise the base \( a \) to get \( b^x \)?". If \( b \) is raised to \( x \), you're essentially counting \( x \) times, leading to the straight multiplication by \( x \).
This rule is particularly handy because it allows us to "pull down" the exponent and convert a more complex exponential expression into a simpler product. This simplification can make subsequent calculations much easier and faster to execute.
Logarithmic Simplification
Simplifying logarithmic expressions can make otherwise complex calculations straightforward. Let's consider the expression \( \log_{4}(4^x) \). Through simplification using properties and rules of logarithms, it becomes much easier to understand and solve.
By applying the Logarithm Power Rule, the initial form \( \log_{4}(4^x) \) reduces to \( x \cdot \log_{4}(4) \). This step highlights that calculating the logarithm of a number raised to a power is equivalent to multiplying that power by the logarithm of the base number.
After applying the rule, recognizing that \( \log_{4}(4) = 1 \) leads to further simplification. The key to simplifying logarithmic expressions is to keep identifying properties and rules that apply to your expression. That helps reduce it to the simplest possible form, making it much easier to handle.
By applying the Logarithm Power Rule, the initial form \( \log_{4}(4^x) \) reduces to \( x \cdot \log_{4}(4) \). This step highlights that calculating the logarithm of a number raised to a power is equivalent to multiplying that power by the logarithm of the base number.
After applying the rule, recognizing that \( \log_{4}(4) = 1 \) leads to further simplification. The key to simplifying logarithmic expressions is to keep identifying properties and rules that apply to your expression. That helps reduce it to the simplest possible form, making it much easier to handle.
Logarithm Base Property
Logarithms have a base property that is a fundamental part of understanding how they work. This property is super straightforward: it's called the Logarithm Base Property. For a logarithm with base \( a \), if you take the logarithm of \( a \) itself, the result is always 1. In notation, this is expressed as \( \log_a(a) = 1 \).
Why is this the case? Logarithms ask the question "what power do we need to raise the base to get a given number?". So, if the number is just the base itself, the answer is simply 1, because \( a^1 = a \).
In practical terms, this property is immensely useful when simplifying expressions. In the expression \( \log_{4}(4^x) \), recognizing that \( \log_{4}(4) \) equals 1 allows us to take an extra step and simplify the exercise down to just \( x \). This kind of simplification showcases the elegance and utility of understanding the deeper properties of logarithms.
Why is this the case? Logarithms ask the question "what power do we need to raise the base to get a given number?". So, if the number is just the base itself, the answer is simply 1, because \( a^1 = a \).
In practical terms, this property is immensely useful when simplifying expressions. In the expression \( \log_{4}(4^x) \), recognizing that \( \log_{4}(4) \) equals 1 allows us to take an extra step and simplify the exercise down to just \( x \). This kind of simplification showcases the elegance and utility of understanding the deeper properties of logarithms.
Other exercises in this chapter
Problem 7
Find the numerical value of the function at the given values of \(a\). $$ g(t)=\sqrt[3]{t} ; a=27,-\frac{1}{8} $$
View solution Problem 7
Let \(f(x)=2 x^{2}+x-4\) and \(g(x)=3-x^{2} .\) Find the specified values. $$ \frac{f(x)-f(2)}{x-2} $$
View solution Problem 8
Solve the equation for \(x\) in \([0,2 \pi)\). $$ \tan x=\sqrt{3} $$
View solution Problem 8
Determine the distance between the given points. \((\sqrt{6}, \sqrt{3})\) and \((3 \sqrt{6},-\sqrt{3})\)
View solution