Problem 54
Question
In \(48-55,\) if \(\log a=c,\) express each of the following in terms of \(c\) $$ \log \left(\frac{a}{10}\right)^{2} $$
Step-by-Step Solution
Verified Answer
\( \log \left(\frac{a}{10}\right)^2 = 2c - 2 \)
1Step 1: Identify the expression
We need to express \( \log \left(\frac{a}{10}\right)^2 \) in terms of \( c \). Since \( \log a = c \), we will express this logarithmic expression using the given variable \( c \).
2Step 2: Use the exponent rule of logarithms
Apply the power rule of logarithms which states: \( \log(b^n) = n\log b \). So, \( \log \left(\frac{a}{10}\right)^{2} = 2 \cdot \log \left(\frac{a}{10}\right) \).
3Step 3: Use the quotient rule of logarithms
Apply the quotient rule, which states: \( \log \left(\frac{x}{y}\right) = \log x - \log y \). Thus, \( \log \left(\frac{a}{10}\right) = \log a - \log 10 \).
4Step 4: Substitute and simplify
Substitute \( \log a = c \) and \( \log 10 = 1 \) (since 10 is the base of the common logarithm system) into the expression. This gives: \( \log \left(\frac{a}{10}\right) = c - 1 \).
5Step 5: Final expression
Substitute \( \log \left(\frac{a}{10}\right) = c - 1 \) back into our expression from step 2: \( 2 \cdot \log \left(\frac{a}{10}\right) = 2(c - 1) = 2c - 2 \). Thus, \( \log \left(\frac{a}{10}\right)^2 = 2c - 2 \).
Key Concepts
Exponent Rule of LogarithmsQuotient Rule of LogarithmsCommon Logarithms
Exponent Rule of Logarithms
Logarithms are powerful tools for simplifying the calculation of exponents. When dealing with an expression like \( \log(b^n) \), the exponent rule of logarithms comes into play. This rule allows us to move the exponent "n" in front of the log function, transforming it to \( n\log(b) \). This simplification is incredibly useful because it turns a more complex expression into a straightforward multiplication problem.
In our exercise, we applied this rule to the expression \( \log \left(\frac{a}{10}\right)^2 \). By bringing the exponent 2 in front, the expression became \( 2 \cdot \log \left(\frac{a}{10}\right) \), making it much simpler to manipulate in subsequent steps.
In our exercise, we applied this rule to the expression \( \log \left(\frac{a}{10}\right)^2 \). By bringing the exponent 2 in front, the expression became \( 2 \cdot \log \left(\frac{a}{10}\right) \), making it much simpler to manipulate in subsequent steps.
Quotient Rule of Logarithms
The quotient rule of logarithms helps divide problems into parts that can be easily simplified. According to this rule, the logarithm of a quotient \( \log \left(\frac{x}{y}\right) \) equals \( \log x - \log y \). This transformation is beneficial to deconstruct complex logarithmic expressions into manageable pieces.
In the problem, we had \( \log \left(\frac{a}{10}\right) \). By applying the quotient rule, this expression breaks down to \( \log a - \log 10 \). This separation is crucial for further simplification, especially when substituting known values from the original exercise context.
In the problem, we had \( \log \left(\frac{a}{10}\right) \). By applying the quotient rule, this expression breaks down to \( \log a - \log 10 \). This separation is crucial for further simplification, especially when substituting known values from the original exercise context.
Common Logarithms
A common logarithm is a logarithm that uses 10 as its base, typically represented as \( \log \) without specifying a base. They simplify calculations in base-10 systems, which is why they are frequently used in mathematical exercises. A key property of common logarithms is that \( \log(10) = 1 \) because ten raised to the first power equals ten itself.
In the step-by-step solution, we utilized this property when substituting known values into the expression. We exchanged \( \log 10 \) with 1 in our equation to help express \( \log \left(\frac{a}{10}\right) \) in terms of "c". This simplification is a perfect showcase of how understanding common logarithms can streamline solving logarithmic problems effectively.
In the step-by-step solution, we utilized this property when substituting known values into the expression. We exchanged \( \log 10 \) with 1 in our equation to help express \( \log \left(\frac{a}{10}\right) \) in terms of "c". This simplification is a perfect showcase of how understanding common logarithms can streamline solving logarithmic problems effectively.
Other exercises in this chapter
Problem 53
Solve each equation for the variable. \(\log _{2} 8+\log _{3} 9=\log _{b} 100,000\)
View solution Problem 54
In \(53-56,\) find each value of \(x\) to the nearest thousandth. $$ e^{x}=217 $$
View solution Problem 54
In \(27-56,\) evaluate each logarithmic expression. Show all work. $$ \frac{2 \log _{1.5} 2.25}{\log _{4} 64-\log _{80} 80} $$
View solution Problem 55
In \(53-56,\) find each value of \(x\) to the nearest thousandth. $$ e^{x}=2 $$
View solution