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Chapter 6: Problem 41

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

Short Answer

Expert verified
The value of the expression is 1.

Step by step solution

01

Understanding the Expression

We need to evaluate \(\log_{2}(3^{-\log_{3}(2)})\). This involves the properties of logarithms and exponents.
02

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 \(\frac{\log_{10}(2)}{\log_{10}(3)}\).
03

Express the Power

The expression inside the logarithm can now be rewritten: \(3^{-\log_{3}(2)} = 3^{-\frac{\log_{10}(2)}{\log_{10}(3)}}\).
04

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.
05

Final Evaluation

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

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Key Concepts

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

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鈥檛 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.

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Most popular questions from this chapter

We list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula \(A(t)=A_{0} e^{k t}\) where \(A_{0}\) is the initial amount of the material and \(k\) is the decay constant. For each isotope: \- Find the decay constant \(k\). Round your answer to four decimal places. \- Find a function which gives the amount of isotope \(A\) which remains after time \(t\). (Keep the units of \(A\) and \(t\) the same as the given data.) \- Determine how long it takes for \(90 \%\) of the material to decay. Round your answer to two decimal places. (HINT: If \(90 \%\) of the material decays, how much is left?) Uranium 235 , used for nuclear power, initial amount \(1 \mathrm{~kg}\) grams, half-life 704 million years.

Use the properties of logarithms to write the expression as a single logarithm. $$ \log _{2}(x)+\log _{4}(x-1) $$

In Exercises \(1-33,\) solve the equation analytically. $$ \frac{150}{1+29 e^{-0.8 t}}=75 $$

For each of the scenarios given in Exercises \(1-6\), \- Find the amount \(A\) in the account as a function of the term of the investment \(t\) in years. \- Determine how much is in the account after 5 years, 10 years, 30 years and 35 years. Round your answers to the nearest cent. \- Determine how long will it take for the initial investment to double. Round your answer to the nearest year. \- Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Round your answer to two decimal places. $$\$ 5000$$ is invested in an account which offers \(2.125 \%\), compounded monthly.

During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in \(1860,\) the Village of Painesville had a population of \(2649 .\) In \(1920,\) the population was 7272\. Use these two data points to fit a model of the form \(N(t)=N_{0} e^{k t}\) were \(N(t)\) is the number of Painesville Residents \(t\) years after \(1860 .\) (Use \(t=0\) to represent the year 1860 . Also, round the value of \(k\) to four decimal places.) According to this model, what was the population of Painesville in \(2010 ?\) (The 2010 census gave the population as 19,563 ) What could be some causes for such a vast discrepancy? For more on this, see Exercise 37 .
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