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Chapter 1: Problem 6

$$ 3^{-\frac{1}{2} \log _{3} 9} \cdot\left\\{\text { Ans. } \frac{1}{3}\right\\} $$

Short Answer

Expert verified
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, \(\frac{1}{3}\).

Step by step solution

01

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.
02

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: \[ 3^{-\frac{1}{2} \cdot 2} \]
03

Simplify the expression further

Now we cancel out 1/2 and 2 in the exponent which gives: \[ 3^{-1} \]
04

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.

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Key Concepts

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

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.
By becoming skilled at using logarithms, such as applying the power rule, you can easily navigate through exercises involving exponential problems.
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.
  • 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.
We then further simplified the exponentiation by recognizing that \(3^{-1}\) means "1 over 3," thereby achieving the simplified expression \(\frac{1}{3}\).
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.
  • 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.
Understanding negative exponents is beneficial not only in simplifying expressions but also in various fields like physics and engineering, where such expressions frequently occur.
With this knowledge, tackling problems involving negative exponents becomes a matter of applying this simple yet powerful rule.

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Most popular questions from this chapter

$$ (\log 2+\log 5+\log 300-\log 3) \cdot 3^{\frac{1}{5 \log _{5} 3}} $$

$$ \left(\log _{\sqrt{5}} \frac{1}{5}\right) \sqrt{\log _{\frac{1}{5}}(5 \sqrt{5})+\log _{\sqrt{5}}(5 \sqrt{5})} $$

$$ \left[\log _{\frac{1}{2}} \sqrt{\frac{1}{4}}+6 \log _{\frac{1}{4}}\left(\frac{1}{2}\right)-2 \log _{\frac{1}{16}}\left(\frac{1}{4}\right)\right] \div \log _{\sqrt{2}} \sqrt[5]{8} $$

$$ \text { If } f(x)=x^{2}-\frac{1}{x^{2}}, \text { prove that } f(x)=-f\left(\frac{1}{x}\right) \text { . } $$

$$ 2^{2-\log _{2} 5} \cdot\left\\{\text { Ans. } \frac{4}{5}\right\\} $$
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