Problem 26
Question
Use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \log _{8}\left(\frac{64}{\sqrt{x+1}}\right) $$
Step-by-Step Solution
Verified Answer
The expanded form of the logarithmic expression is \(2 - \frac{1}{2}\log _{8}(x+1)\).
1Step 1: Apply the Quotient Rule
The quotient rule of logarithms states that \(\log_b(m/n) = \log_b(m) - \log_b(n)\). Applying this rule, the equation \(\log _{8}\left(\frac{64}{\sqrt{x+1}}\right)\) is expanded as \(\log _{8}(64) - \log _{8}(\sqrt{x+1})\).
2Step 2: Evaluate the Logarithm Base 8 of 64 and Apply the Power Rule
The logarithm base 8 of 64 is evaluated as 2 because 8 raised to the power of 2 is 64. For the second half of the equation, the power rule of logarithms states that \(\log_b(m^n) = n\cdot \log_b(m)\). Applying this rule and remembering that the square root is the same as raising to the power of \(\frac{1}{2}\), the equation is now \(2 - \frac{1}{2}\log _{8}(x+1)\).
3Step 3: Final Answer
The final answer to the expanded logarithm equation is \(2 - \frac{1}{2}\log _{8}(x+1)\). It cannot be further evaluated without the value of x.
Key Concepts
Quotient Rule of LogarithmsPower Rule of LogarithmsEvaluating Logarithms
Quotient Rule of Logarithms
The quotient rule of logarithms is a fundamental property that enables us to simplify the logarithm of a quotient, or division, between two numbers. When faced with a logarithmic expression like \( \log _{b}\left(\frac{m}{n}\right) \), where \( b \) is the base, \( m \) is the numerator, and \( n \) is the denominator, the quotient rule allows us to separate this single log into the difference of two logs, each with the same base. Mathematically, this rule is written as \( \log_b(m/n) = \log_b(m) - \log_b(n) \).
This property comes in very handy when you're trying to simplify complex logarithmic expressions or when evaluating logarithms without a calculator, as you can break down complex terms into more manageable pieces. For example, if you have \( \log _{8}\left(\frac{64}{\sqrt{x+1}}\right) \), as encountered in the exercise, you can apply this rule to break it into \( \log _{8}(64) - \log _{8}(\sqrt{x+1}) \), which is much simpler to work with. It also opens the door to using other logarithmic properties, such as the power rule, to further simplify or calculate the logarithms.
This property comes in very handy when you're trying to simplify complex logarithmic expressions or when evaluating logarithms without a calculator, as you can break down complex terms into more manageable pieces. For example, if you have \( \log _{8}\left(\frac{64}{\sqrt{x+1}}\right) \), as encountered in the exercise, you can apply this rule to break it into \( \log _{8}(64) - \log _{8}(\sqrt{x+1}) \), which is much simpler to work with. It also opens the door to using other logarithmic properties, such as the power rule, to further simplify or calculate the logarithms.
Power Rule of Logarithms
The power rule of logarithms turns a cumbersome expression into a more workable one by dealing with exponents within the logarithm. Stated simply, if you have an expression such as \( \log_b(m^n) \), where \( m \) is the base of the logarithm, \( n \) is the exponent, and \( b \) remains as the base of the logarithm, you can rewrite this as \( n\cdot \log_b(m) \).
This rule is particularly useful when encountering logarithmic expressions containing roots or higher powers, which can be challenging to evaluate without translation. In the textbook example, the power rule is applied to \( \log _{8}(\sqrt{x+1}) \) by recognizing that \( \sqrt{x+1} \) is the same as raising \( x+1 \) to the power of \( \frac{1}{2} \). Thus, the power rule lets us rewrite this part of the expression as \( \frac{1}{2}\log _{8}(x+1) \), making it clearer and easier to handle if there's further simplification or evaluation to be performed.
This rule is particularly useful when encountering logarithmic expressions containing roots or higher powers, which can be challenging to evaluate without translation. In the textbook example, the power rule is applied to \( \log _{8}(\sqrt{x+1}) \) by recognizing that \( \sqrt{x+1} \) is the same as raising \( x+1 \) to the power of \( \frac{1}{2} \). Thus, the power rule lets us rewrite this part of the expression as \( \frac{1}{2}\log _{8}(x+1) \), making it clearer and easier to handle if there's further simplification or evaluation to be performed.
Evaluating Logarithms
Evaluating logarithms involves finding the exponent that the base must be raised to in order to produce the number inside the logarithm. This process can sometimes be straightforward, as in evaluating \( \log _{8}(64) \), which is a matter of asking ourselves '8 to what power gives us 64?'. In this case, the answer is 2 since \( 8^2 = 64 \).
However, when the logarithm cannot be easily translated into a simple exponentiation, we may need to rely on logarithmic properties, as previously mentioned, or on intuitive understanding or even calculators for approximation. Remember, logarithms are the inverses of exponentiation, so evaluating them is essentially reversing the process of raising a number to a power. When it's not possible to evaluate a logarithm exactly, such as in expressions involving variables like \( \log _{8}(x+1) \), it's typically left in the logarithmic form until further information (like a specific value for \( x \)) is provided.
In educational contexts, the goal is to ensure that students can manipulate and understand logarithmic expressions, even if they can't always find a numerical answer. The step-by-step example provided showcases this approach, moving from raw expressions to an evaluated form where possible and leaving the rest in its logarithmic state.
However, when the logarithm cannot be easily translated into a simple exponentiation, we may need to rely on logarithmic properties, as previously mentioned, or on intuitive understanding or even calculators for approximation. Remember, logarithms are the inverses of exponentiation, so evaluating them is essentially reversing the process of raising a number to a power. When it's not possible to evaluate a logarithm exactly, such as in expressions involving variables like \( \log _{8}(x+1) \), it's typically left in the logarithmic form until further information (like a specific value for \( x \)) is provided.
In educational contexts, the goal is to ensure that students can manipulate and understand logarithmic expressions, even if they can't always find a numerical answer. The step-by-step example provided showcases this approach, moving from raw expressions to an evaluated form where possible and leaving the rest in its logarithmic state.
Other exercises in this chapter
Problem 25
In Exercises 21–42, evaluate each expression without using a calculator. $$ \log _{5} \frac{1}{5} $$
View solution Problem 25
begin by graphing \(f(x)=2^{x} .\) Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes Use
View solution Problem 26
Solve each exponential equation . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approx
View solution Problem 26
In Exercises 21–42, evaluate each expression without using a calculator. $$ \log _{6} \frac{1}{6} $$
View solution