Problem 40
Question
Approximate each logarithm to three decimal places. $$ \log _{5} 0.047 $$
Step-by-Step Solution
Verified Answer
\( \log_{5} 0.047 \approx -1.898 \).
1Step 1: Apply Change of Base Formula
To compute \( \log_{5} 0.047 \), use the change of base formula: \( \log_{a} b = \frac{\log_{10} b}{\log_{10} a} \). Therefore, \( \log_{5} 0.047 = \frac{\log_{10} 0.047}{\log_{10} 5} \).
2Step 2: Compute \( \log_{10} 0.047 \)
Calculate \( \log_{10} 0.047 \) using a calculator. It is approximately \( -1.327 \) (rounded to three decimal places).
3Step 3: Compute \( \log_{10} 5 \)
Calculate \( \log_{10} 5 \) using a calculator. It is approximately \( 0.699 \) (rounded to three decimal places).
4Step 4: Divide the Results
Apply the division from Step 1: \( \frac{-1.327}{0.699} = -1.898 \).
5Step 5: Approximated Result
Upon calculating and rounding to three decimal places, \( \log_{5} 0.047 \approx -1.898 \).
Key Concepts
Change of Base FormulaLogarithmic ApproximationCommon Logarithms
Change of Base Formula
The change of base formula is a powerful tool when working with logarithms. Often, we deal with uncommon or difficult bases, like base 5 in the example problem. This formula allows us to convert a logarithm from one base to another, often making it easier to solve.
The formula is expressed as:
In the formula, \(a\) and \(b\) are the original numbers in the logarithm calculation, and you are transforming it to use the base 10 logarithms. This makes the change of base formula extremely useful for simplifying calculations in mathematics.
The formula is expressed as:
- \( \log_{a} b = \frac{\log_{10} b}{\log_{10} a} \)
In the formula, \(a\) and \(b\) are the original numbers in the logarithm calculation, and you are transforming it to use the base 10 logarithms. This makes the change of base formula extremely useful for simplifying calculations in mathematics.
Logarithmic Approximation
Logarithmic approximation is crucial when solving logarithms without exact solutions or when using tools that provide decimal places, like calculators. Often, we're required to find the logarithm of an exact value but only need it to a specific number of decimal places.
Take the provided problem where \( \log_{10} 0.047 \) is calculated and approximated to \(-1.327\). This is achieved using a calculator, which gives a precise value that can be approximated further. The need to approximate is common in mathematics as not all logarithmic values can be expressed neatly or with a finite number of digits.
In practice, this approximation step helps to produce results that are good enough for practical purposes, such as scientific calculations or real-world applications where being precise is less critical than being close to the true value.
Take the provided problem where \( \log_{10} 0.047 \) is calculated and approximated to \(-1.327\). This is achieved using a calculator, which gives a precise value that can be approximated further. The need to approximate is common in mathematics as not all logarithmic values can be expressed neatly or with a finite number of digits.
In practice, this approximation step helps to produce results that are good enough for practical purposes, such as scientific calculations or real-world applications where being precise is less critical than being close to the true value.
Common Logarithms
Common logarithms, or base 10 logarithms, are a standard form in mathematical computations. They are denoted as \( \log_{10} \) or even just \( \log \) when the base is not specified. These logarithms are fundamental because of their simplicity and ease of use with calculators, which almost always support base 10 logarithms directly.
In simpler terms, a common logarithm answers the question: "To what power must 10 be raised to produce a given number?" For instance, \( \log_{10} 100 = 2 \) because \(10^2 = 100\).
The advantage of common logarithms comes from their universality in scientific notation, where powers of 10 are frequent, and in financial calculations, simplifying complex exponential growth or decay problems. They are also integral to the change of base formula, as seen in converting \( \log_{5} 0.047 \) into a common logarithm calculation to ease the solution process.
In simpler terms, a common logarithm answers the question: "To what power must 10 be raised to produce a given number?" For instance, \( \log_{10} 100 = 2 \) because \(10^2 = 100\).
The advantage of common logarithms comes from their universality in scientific notation, where powers of 10 are frequent, and in financial calculations, simplifying complex exponential growth or decay problems. They are also integral to the change of base formula, as seen in converting \( \log_{5} 0.047 \) into a common logarithm calculation to ease the solution process.
Other exercises in this chapter
Problem 39
For Problems \(39-44\), graph each of the exponential functions. See answer section. $$ f(x)=e^{x}+1 $$
View solution Problem 39
Graph each of the exponential functions. $$ f(x)=2^{-x-2} $$
View solution Problem 40
Use your calculator to find \(x\) when given \(\ln x\). Express answers to five significant digits. $$ \ln x=-2.3745 $$
View solution Problem 40
Evaluate each logarithmic expression. $$ \log _{2}\left(\log _{5} 5\right) $$
View solution