Problem 44

Question

\(\operatorname{In} 36-47,\) if \(\log 3=x\) and \(\log 5=y,\) write each of the logs in terms of \(x\) and \(y\) $$ \log 0.04 $$

Step-by-Step Solution

Verified
Answer
\( \log 0.04 = -1 - 2y \)
1Step 1: Express 0.04 as a Power of 10
Recognize that 0.04 can be written as a power of 10: \( 0.04 = \frac{4}{100} = \frac{2^2}{10^2} = \frac{2^2}{10^2} = 4 \times 10^{-2} \).
2Step 2: Use the Logarithmic Identity
Apply the property of logarithms that states \( \log(ab) = \log(a) + \log(b) \). For \( \log(4 \times 10^{-2}) \), we have \( \log 4 + \log 10^{-2} \).
3Step 3: Simplify \( \log 10^{-2} \)
Using the logarithm property \( \log(b^n) = n \cdot \log(b) \), calculate \( \log 10^{-2} = -2 \cdot \log 10 \). Since \( \log 10 = 1 \), it follows that \( \log 10^{-2} = -2 \).
4Step 4: Find \( \log 4 \) in terms of \( x \) and \( y \)
Express \( \log 4 \) using \( \log(a^n) = n \cdot \log(a) \). Recognize that \( 4 = 2^2 \) and \( 2 = \sqrt{10}/5 = 10^{1/2}/5 \). Using the given \( \log 5 = y \), we find \( \log 2 = \frac{1}{2} - y \). Thus, \( \log 4 = 2 \cdot (\frac{1}{2} - y) = 1 - 2y \).
5Step 5: Combine Results
Combine the results from the previous steps: \( \log 0.04 = \log 4 + \log 10^{-2} = (1 - 2y) - 2 = -1 - 2y \).

Key Concepts

Logarithmic IdentitiesExpressing Logs in Terms of VariablesSimplifying Logarithmic Expressions
Logarithmic Identities
Logarithmic identities are mathematical rules which help to simplify the expression and manipulation of logarithms. Understanding these identities can make complex problems much simpler to solve.
One fundamental identity is the product property of logarithms, which states: \( \log(ab) = \log(a) + \log(b) \). This property allows us to break down the logarithm of a product into the sum of individual logarithms. Another key identity is the power property: \( \log(b^n) = n \cdot \log(b) \). This lets us take the exponent out as a multiplier in front of the logarithm.
These properties are pivotal when tackling problems involving the transformation and simplification of expressions with logarithms. By using identities, we can express complex logarithmic expressions in simpler terms, allowing for easier calculation or further manipulation.
Expressing Logs in Terms of Variables
Expressing logs in terms of variables involves rewriting logarithmic expressions using given variables. This is especially helpful when dealing with specific values of logarithms given as variables, like \( \log 3 = x \) and \( \log 5 = y \).
For example, if we need to express \( \log 4 \) in terms of \( x \) and \( y \), we can use the power identity \( \log(2^2) = 2 \cdot \log(2) \). First, express 2 using the variables: since 2 can be considered in terms of earlier given variables like \( \log 5 \), find a way to manipulate those expressions effectively (involving other relatable values such as \( 10 \)).
  • Utilize identities to find intermediate expressions
  • Substitute known values where possible
  • Simplify step by step to reach the desired expression
By expressing these logs in terms of known variables, we create new pathways to solve or simplify problems further.
Simplifying Logarithmic Expressions
Simplifying logarithmic expressions often involves using the previously mentioned identities to reduce the complexity of the equation. The goal is to express the logarithm in its simplest form for easy interpretation or computation.
Let's consider an example where we aim to simplify \( \log 0.04 \). Begin by expressing 0.04 as a power of 10, such as \( 4 \times 10^{-2} \). Using the product property, separate this into \( \log 4 + \log 10^{-2} \).
Next, apply individual logarithmic identities:
  • For \( \log 10^{-2} \), use the power property to get \( -2 \cdot \log 10 \). Since \( \log 10 = 1 \), this simplifies to \( -2 \).
  • For \( \log 4 \), use existing expressions, like replacing it with \( 1 - 2y \), where the computations have determined that using other known values.
Combining these simplified components leads us to a final expression which is manageable and effective for calculations, showing how identities streamline logarithmic problem-solving.