Question

a) Prove the change of base formula, \(\log _{e} x=\frac{\log _{d} x}{\log _{d} c},\) where \(c\) and \(d\) are positive real numbers other than \(1 .\) b) Apply the change of base formula for base \(d=10\) to find the approximate value of \(\log _{2} 9.5\) using common logarithms. Answer to four decimal places. c) The Krumbein phi ( \(\varphi\) ) scale is used in geology to classify the particle size of natural sediments such as sand and gravel. The formula for the \(\varphi\) -value may be expressed as \(\varphi=-\log _{2} D,\) where \(D\) is the diameter of the particle, in millimetres. The \(\varphi\) -value can also be defined using a common logarithm. Express the formula for the \(\varphi\) -value as a common logarithm. d) How many times the diameter of medium sand with a \(\varphi\) -value of 2 is the diameter of a pebble with a \(\varphi\) -value of \(-5.7 ?\) Determine the answer using both versions of the \(\varphi\) -value formula from part c).

Short answer

a) Proven: \(\text{log}_{e} x = \frac{\text{log}_{d} x}{\text{log}_{d} c}\). b) \(\text{log}_{2} 9.5 \approx 3.2488\). c) \(\text{Phi} = -\frac{\text{log}_{10} D}{\text{log}_{10} 2}\). d) Medium sand's diameter is about 213.90 times greater than the pebble's.

Step-by-step solution

Prove the change of base formula

To prove \(\text{log}_{e} x = \frac{\text{log}_{d} x}{\text{log}_{d} c}\), where c and d are positive real numbers other than 1:Consider \(\text{log}_{d} x\) and rewrite it using the definition of logarithms: \(d^{\text{log}_{d} x} = x\). Now take the natural logarithm (base e) of both sides: \(\text{ln}(d^{\text{log}_{d} x}) = \text{ln}(x)\). Using the power rule for logarithms \(\text{ln}(a^b) = b \text{ln}(a)\): \(\text{log}_{d} x \text{ln}(d) = \text{ln}(x)\). Isolate \(\text{log}_{d} x\) by dividing both sides by \(\text{ln}(d)\): \(\text{log}_{d} x = \frac{\text{ln}(x)}{\text{ln}(d)}\). Replacing natural logarithm with logarithm base c: \(\text{log}_{e} x = \frac{\text{log}_{d} x}{\text{log}_{d} c}\), proving the change of base formula.

Apply change of base formula to find \(\text{log}_{2} 9.5\)

Given d = 10, apply the formula \(\text{log}_{2} 9.5 = \frac{\text{log}_{10} 9.5}{\text{log}_{10} 2}\). Use a calculator to find \(\text{log}_{10} 9.5\) and \(\text{log}_{10} 2\): \(\text{log}_{10} 9.5 \approx 0.9777\) and \(\text{log}_{10} 2 \approx 0.3010\). Calculate \(\frac{0.9777}{0.3010} \approx 3.2488\). Therefore, \(\text{log}_{2} 9.5 \approx 3.2488\).

Express \(\text{log}_{2} D \) in common logarithms

Given \(\text{Phi} = -\text{log}_{2} D\). Use the change of base formula to write \(\text{log}_{2} D\) in terms of common logarithms: \(\text{log}_{2} D = \frac{\text{log}_{10} D}{\text{log}_{10} 2}\). Therefore, \(\text{Phi} = -\frac{\text{log}_{10} D}{\text{log}_{10} 2}\).

Calculate \(D\text{ values for given }\text{Phi}\text{ - values}\)

For \(\text{Phi} = 2\), use the formula \(\text{Phi} = -\text{log}_{2} D\): \(-2 = -\text{log}_{2} D\), implying \(\text{log}_{2} D = 2\). Calculate \(D = 2^2 = 4\) mm.For \(\text{Phi} = -5.7\), \(-(-5.7) = -\text{log}_{2} D\), implying \(\text{log}_{2} D = -5.7\). Calculate \(D = 2^{-5.7} \approx 0.0187\) mm.

Compare diameters

The diameter of medium sand \(D_{\text{sand}} = 4\) mm, and the diameter of a pebble \(D_{\text{pebble}} = 0.0187\) mm. Therefore, the ratio of the diameters is \(\frac{4}{0.0187} \approx 213.90\). The diameter of a medium sand particle is approximately 213.90 times the diameter of the pebble.

Key concepts

logarithms

A logarithm is a mathematical function that helps us understand how one number can be expressed as a power of another number. Essentially, if you have an equation of the form \(a^b = c\) then \(b\) is the logarithm of \(c\) base \(a\), expressed as \( \text{log}_{a}(c) = b\). Logarithms are widely used in various fields such as science, engineering, and finance to make complex multiplications and divisions more manageable. They help convert multiplicative relationships into additive ones by leveraging the properties of exponents. For example, \(\text{log}_a(x \times y) = \text{log}_a(x) + \text{log}_a(y)\).

When dealing with logarithms, it is crucial to understand their rules and properties:

• Product Rule: \(\text{log}_a(x \times y) = \text{log}_a(x) + \text{log}_a(y)\)
• Quotient Rule: \(\text{log}_a(\frac{x}{y}) = \text{log}_a(x) - \text{log}_a(y)\)
• Power Rule: \(\text{log}_a(x^b) = b \times \text{log}_a(x)\)
• Change of Base Formula: \(\text{log}_a(x) = \frac{\text{log}_d(x)}{\text{log}_d(a)}\), where \(d\) is any positive real number other than 1.

These properties can simplify complex calculations involving exponential growth and decay.

common logarithms

Common logarithms are logarithms with a base of 10, denoted \( \text{log}_{10} \) or more simply as \( \text{log}\). They are often used in scientific and engineering calculations because our numerical system is base 10. When dealing with common logarithms, you can use the change of base formula to convert logarithms of other bases. For example, to find \( \text{log}_2(9.5)\), you can use the common logarithm:

• Apply: \( \text{log}_2(9.5) = \frac{\text{log}_{10}(9.5)}{\text{log}_{10}(2)}\)
• Use a calculator to find \( \text{log}_{10}(9.5) \text{ and } \text{log}_{10}(2)\): \( \text{log}_{10}(9.5) \text{ ≈ } 0.9777\) and \( \text{log}_{10}(2) \text{ ≈ } 0.3010\)
• Perform the division: \( \frac{0.9777}{0.3010} \text{ ≈ } 3.2488\)

Thus, \( \text{log}_2(9.5) \text{ ≈ } 3.2488\). Common logarithms simplify many scientific calculations involving growth rates, sound intensity, and even the Richter scale for measuring earthquakes.

phi value

The phi value \( (\text{ϕ}) \) is a measurement used in geology to determine the particle size of sediments such as sand and gravel. The formula for calculating the phi-value based on the particle diameter \( (D) \) in millimeters uses a logarithmic scale: \( \text{ϕ} = -\text{log}_2(D)\). This categorical system provides a way to standardize and classify particle sizes for various geological applications. If you want to express the phi value in terms of common logarithms, you can adjust the formula using the change of base rule:

• First, write the original formula: \( \text{ϕ} = -\text{log}_2(D)\)
• Apply the change of base formula for logarithms: \( \text{log}_2(D) = \frac{\text{log}_{10}(D)}{\text{log}_{10}(2)}\)
• Substitute this into the phi formula: \( \text{ϕ} = -\frac{\text{log}_{10}(D)}{\text{log}_{10}(2)}\)

This conversion allows for easy use of scientific calculators that primarily support common logarithms. Utilizing the phi value in this form provides versatile and precise classifications for sediment analysis.

particle size classification

Particle size classification is a crucial part of sedimentology — the study of sediments, including sand, silt, clay, and gravel. Using the Krumbein phi scale, particles are classified based on their diameters measured in millimeters. Understanding and calculating these classifications help geologists and engineers assess and manage natural sediments effectively. Using phi values makes it easier to compare sizes. Here’s how you can calculate and differentiate between sizes using phi values:

• Suppose we have a medium sand particle with a phi value of 2.
• Using the formula \( \text{ϕ} = -\text{log}_2(D)\), we solve for \( D \), where \( \text{ϕ} = 2\):
• \( -2 = -\text{log}_2(D)\)\, so \( D = 2^2 = 4\) mm.
• For a pebble with a phi value of -5.7:
• \( -(-5.7) = -\text{log}_2(D)\)\, so \( \text{log}_2(D) = -5.7\):
• \( D = 2^{-5.7} \text{ ≈ } 0.0187\) mm.

Thus, the medium sand particle's diameter is \( 4\) mm, and the pebble's diameter is approximately \( 0.0187\) mm. To compare, calculate the ratio of their diameters: \( \frac{D_{\text{sand}}}{D_{\text{pebble}}} \text{ ≈ } \frac{4}{0.0187} \text{ ≈ } 213.90\). This example shows that the diameter of medium sand is approximately 213.90 times that of the pebble.