Problem 6
Question
$$ 3^{-\frac{1}{2} \log _{3} 9} \cdot\left\\{\text { Ans. } \frac{1}{3}\right\\} $$
Step-by-Step Solution
Verified Answer
The expression \(3^{-\frac{1}{2} \log _{3} 9}\) simplifies to \(3^{-1}\) after simplifying the logarithm using power rule and cancelling out the 1/2 and 2 in the exponent. Evaluating \(3^{-1}\) gives the reciprocal of 3 as the final answer, \(\frac{1}{3}\).
1Step 1: Understand the expression
The given exercise is \(3^{-\frac{1}{2} \log _{3} 9}\). Here \(\log _{3} 9\) is the logarithm of 9 to the base 3. It is raised to the power -1/2 and then used as the exponent of 3.
2Step 2: Simplify the logarithm
This step involves simplifying the expression \(\log _{3} 9\). This could be done using logarithm power rule which states:
\[
\log_b {a^n} = n \log_b {a}
\]
We can rewrite 9 as \(3^2\). Replacing this in the original expression gives:
\[
3^{-\frac{1}{2} \log _{3} (3^2)}
\]
Using the power rule of logarithms the expression simplifies to:
\[
3^{-\frac{1}{2} \cdot 2}
\]
3Step 3: Simplify the expression further
Now we cancel out 1/2 and 2 in the exponent which gives:
\[
3^{-1}
\]
4Step 4: Evaluate \(3^{-1}\)
The expression \(3^{-1}\) is the reciprocal of 3 because negative exponents refer to the reciprocal of the positive exponent. Hence, \(3^{-1}= \frac{1}{3}\).
So, the given expression \(3^{-\frac{1}{2} \log _{3} 9}\) simplifies to \(\frac{1}{3}\). This corresponds with the provided answer. Hence, the given expression has been correctly simplified.
Key Concepts
LogarithmsSimplifying expressionsNegative exponents
Logarithms
Logarithms are a fundamental concept in mathematics that help us solve equations involving exponents. Understanding logarithms can greatly simplify complex problems. Let's break it down:
- Logarithms determine how many times a base number needs to be multiplied to reach another number. For example, \(\log_3 9\) asks us, "3 raised to what power equals 9?".
- In this case, 9 can be expressed as \(3^2\), so \(\log_3 9 = 2\).
- Knowing how to manipulate logarithms is crucial. The power rule of logarithms, which states \(\log_b {a^n} = n \log_b {a}\), allows us to bring the exponent in front of the logarithm, making it easier to simplify expressions.
Simplifying expressions
Simplifying expressions is an essential skill in mathematics that involves reducing a mathematical expression to its simplest form. This usually means making it shorter or more straightforward while retaining its original value.
Simplifying isn't just about eliminating complexity; it's about making an expression easier to understand and solve.
- In the context of our problem, we first needed to handle \(\log_3 9\) using the identity 9 equals \(3^2\) to rewrite the expression.
- Applying the logarithmic power rule, the expression "\(-\frac{1}{2} \log_3(3^2)\)" was simplified to \(-1\), since \(\frac{1}{2}\) and \(2\) cancel each other out.
Simplifying isn't just about eliminating complexity; it's about making an expression easier to understand and solve.
Negative exponents
Negative exponents might seem confusing, but they can be easily understood with a simple rule in mathematics: a negative exponent indicates the reciprocal.
With this knowledge, tackling problems involving negative exponents becomes a matter of applying this simple yet powerful rule.
- If you have a negative exponent, such as \(a^{-b}\), it means you need to take the reciprocal of "a" raised to the positive power "b." In mathematical terms, it becomes \(\frac{1}{a^b}\).
- Applying this rule to our problem: \(3^{-1}\) becomes \(\frac{1}{3}\).
- The negative exponent essentially "flips" the base fraction, offering another way to express division.
With this knowledge, tackling problems involving negative exponents becomes a matter of applying this simple yet powerful rule.
Other exercises in this chapter
Problem 4
$$ \begin{aligned} &\text { If } \begin{aligned} f(x) &=\frac{x^{2}+1}{x-1}, \quad x3, \end{aligned}\\\ &\text { for what values of } x \text { is the function
View solution Problem 5
$$ \begin{aligned} &\text { Are the following functions equivalent? }\\\ &\text { i. } \left.\quad f(x)=\frac{x}{x} \text { and } \phi(x)=1 \text { \\{Ans. No }
View solution Problem 7
$$ 2^{2-\log _{2} 5} \cdot\left\\{\text { Ans. } \frac{4}{5}\right\\} $$
View solution Problem 8
$$ 10^{\log m+\log n} . $$
View solution