Problem 41 - Precalculus

Question

Evaluate the expression. \(\log _{2}\left(3^{-\log _{3}(2)}\right)\)

Step-by-Step Solution

Answer

The value of the expression is 1.

Step 1: Understanding the Expression

We need to evaluate \(\log_{2}(3^{-\log_{3}(2)})\). This involves the properties of logarithms and exponents.

Step 2: Applying the Change of Base Formula

First, use the change of base formula \(\log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)}\) for the inner logarithm, \(\log_{3}(2)\). This can be rewritten as \(\frac{\log_{10}(2)}{\log_{10}(3)}\).

Step 3: Express the Power

The expression inside the logarithm can now be rewritten: \(3^{-\log_{3}(2)} = 3^{-\frac{\log_{10}(2)}{\log_{10}(3)}}\).

Step 4: Using Properties of Logarithms

Apply the properties of exponents. \(3^{-\frac{\log_{10}(2)}{\log_{10}(3)}}\) is equivalent to \(\left(3^{\log_{10}(3)}\right)^{-\log_{10}(2)}\). This simplifies using exponent rules to \(10^{\log_{10}(2)}\), which equals 2.

Step 5: Final Evaluation

Substituting back into the original expression, we now have \(\log_{2}(2)\). Since \(\log_{b}(b) = 1\), it simplifies directly to 1.

Key Concepts

Change of Base Formula

Logarithms allow us to go between exponential and linear scales. Sometimes you're given a logarithm with a base that's not easy to work with. That's where the change of base formula helps. It lets us rewrite a logarithm in terms of two other bases. The formula is: \[\log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)}\]This is especially handy when using calculators or when simplifying complex logarithmic expressions. In our example, we transformed the base of the logarithm from 3 to 10, a common choice because many calculators primarily support base 10 (common logarithm) or base \(e\) (natural logarithm). By applying this formula, internal calculations become manageable.

• Easy to use when calculators don’t have specific base functions.
• Simplifies complex logarithmic expressions.
• Makes operations with non-standard bases more straightforward.

Properties of Exponents

Understanding the properties of exponents is crucial when working with logarithmic expressions. Exponents help express repeated multiplication. Several key properties aid in simplifying expressions:

• Power of a Product: \((mn)^a = m^a \times n^a\)
• Power of a Power: \((m^a)^b = m^{a\times b}\)
• Zero Exponent: \(m^0 = 1\) for any non-zero \(m\)
• Negative Exponent: \(m^{-a} = \frac{1}{m^a}\)

In our exercise, we modified an exponential expression through these properties. Transitioning from \(3^{-\frac{\log_{10}(2)}{\log_{10}(3)}}\) to simpler forms showcases how properties allow exponents and logarithms to simplify neatly to direct values, like 2.

Properties of Logarithms

Logarithms have several properties that make manipulating them easier, quite similar to how we handle exponents. These include:

• Product Rule: \(\log_{b}(mn) = \log_{b}(m) + \log_{b}(n)\)
• Quotient Rule: \(\log_{b}\left(\frac{m}{n}\right) = \log_{b}(m) - \log_{b}(n)\)
• Power Rule: \(\log_{b}(m^a) = a\cdot\log_{b}(m)\)
• Base Switch: \(\log_{b}(b) = 1\)

In the original exercise, after applying the exponent properties, reaching \(\log_{2}(2)\) leverages the base switch property. Since any log of a number in its own base equals one, this simplifying property resulted in the final answer. Utilizing these properties effectively helps decode complex logarithmic problems into digestible parts, as seen in this exercise.