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Chapter 6: Problem 28

Evaluate the expression. \(\log _{4}(8)\)

Short Answer

Expert verified
The value of \( \log _{4}(8) \) is \( \frac{3}{2} \).

Step by step solution

01

Understanding the Logarithmic Expression

We need to evaluate \( \log_{4}(8) \). This expression asks what power we should raise the base, 4, to get 8.
02

Change of Base Formula

When computing logarithms, it's often useful to use the change of base formula: \( \log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)} \). We'll use base 2 because both 4 and 8 are powers of 2.
03

Convert to Base 2

Convert the expression using base 2: \( \log_{4}(8) = \frac{\log_{2}(8)}{\log_{2}(4)} \).
04

Evaluate \(\log_{2}(8)\) and \(\log_{2}(4)\)

Calculate \( \log_{2}(8) \): since \( 8 = 2^3 \), \( \log_{2}(8) = 3 \). Calculate \( \log_{2}(4) \): since \( 4 = 2^2 \), \( \log_{2}(4) = 2 \).
05

Final Calculation

Substitute these values back into the converted expression: \( \log_{4}(8) = \frac{3}{2} \).

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Key Concepts

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

Understanding the Change of Base Formula
The change of base formula is a very useful tool when dealing with logarithms. It allows us to rewrite a logarithm in a different base, which can make calculations much easier. The formula is given as: \( \log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)} \). This means you can convert a logarithm with base \( b \) into a fraction of two logarithms with a new base \( c \).
  • This formula is particularly helpful when dealing with bases that are not common or easy to work with.
  • By using a base that makes calculations simpler, such as base 2 or 10, you can evaluate the original expression more easily.
In the given problem, we used the change of base formula to convert \( \log_{4}(8) \) into \( \frac{\log_{2}(8)}{\log_{2}(4)} \) because both 4 and 8 are powers of 2.
This strategy simplifies the task significantly, as calculating powers of 2 is straightforward.
Breaking Down Logarithmic Expressions
Logarithmic expressions can often seem intimidating, but breaking them down makes them easier to understand. A logarithmic expression, such as \( \log_{b}(a) \), asks the question: "To what power must we raise the base \( b \) to obtain the number \( a \)?"
This means if you have \( \log_{b}(a) = x \), \( b^x = a \).
  • Understanding that logarithms are the inverse operation of exponentiation is key. Just as subtraction is the inverse of addition, logarithms undo exponentiation.
  • In our example, \( \log_{4}(8) \) asks, "What power do we raise 4 to, in order to get 8?"
By using properties of logarithms and transformations like the change of base formula, these questions become manageable operations. Breaking complex problems into simpler parts often involves transforming the base and using properties of exponents.
The Role of Exponents in Logarithms
Exponents play a central role in understanding and evaluating logarithms. When evaluating a logarithm, you are essentially finding an exponent. If \( b^x = a \), it follows that \( \log_{b}(a) = x \).
  • A strong understanding of exponent rules simplifies the process of working with logarithms.
  • In our exercise, understanding that \( 8 = 2^3 \) and \( 4 = 2^2 \) helped us evaluate the logarithm using base 2.
Using the rule that \( b^x = a \), to find \( \log_{b}(a) \), you essentially find which exponent \( x \) makes the equation true.
This relationship between exponents and logarithms turns complex expressions into manageable challenges. Whenever you see a logarithm, think of it in terms of its related exponent. This mental conversion makes logarithmic expressions less daunting to tackle.

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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?) Phosphorus 32 , used in agriculture, initial amount 2 milligrams, half-life 14 days.

Find the domain of the function. \(f(x)=\log _{7}(4 x+8)\)

Sketch the graph of \(y=g(x)\) by starting with the graph of \(y=f(x)\) and using transformations. Track at least three points of your choice and the horizontal asymptote through the transformations. State the domain and range of \(g\). . \(f(x)=10^{x}, g(x)=10^{\frac{x+1}{2}}-20\)

We introduce three widely used measurement scales which involve common logarithms: the Richter scale, the decibel scale and the pH scale. The computations involved in all three scales are nearly identical so pay attention to the subtle differences. While the decibel scale can be used in many disciplines, \(^{13}\) we shall restrict our attention to its use in acoustics, specifically its use in measuring the intensity level of sound. \(^{14}\) The Sound Intensity Level \(L\) (measured in decibels) of a sound intensity \(I\) (measured in watts per square meter) is given by $$L(I)=10 \log \left(\frac{I}{10^{-12}}\right)$$ Like the Richter scale, this scale compares \(I\) to baseline: \(10^{-12} \frac{W}{m^{2}}\) is the threshold of human hearing. (a) Compute \(L\left(10^{-6}\right)\). (b) Damage to your hearing can start with short term exposure to sound levels around 115 decibels. What intensity \(I\) is needed to produce this level? (c) Compute \(L(1)\). How does this compare with the threshold of pain which is around 140 decibels?

Solve the equation analytically. $$ \log (x)-\log (2)=\log (x+8)-\log (x+2) $$
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