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The average rate of change measures how a function's value changes on average over a specific interval. Think of it as a way to measure the slope of the function over that interval. The formula for the average rate of change of a function \( f(x) \) over an interval \([a, b]\) is:

• \( \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \)

In our exercise, you have \( f(x) = \log_{10}(x) \). To find the average rate of change between \( x=1 \) and \( x=3 \), we used the values from our table:

• \( f(1) = 0.00 \)
• \( f(3) = 0.48 \)

By plugging these values into our formula, we calculated:

• \( \frac{0.48 - 0.00}{3-1} = 0.24 \)

This means that on average, our logarithmic function increases by 0.24 units per unit increase in \( x \) over the interval from 1 to 3. It's like finding the straight-line path of change between two points on a curve.

The instantaneous rate of change is a little more complex as it involves understanding how a function behaves at an exact moment rather than over an interval. It corresponds to the slope of the tangent line at a given point on a curve, offering a snapshot of how the function is changing at just that point.
To approximate this for \( f(x) = \log_{10}(x) \) at \( x=2 \), we used the average rate of change over a small interval around \( x=2 \), specifically \([1.5, 2.5]\). This method provides a close estimation by balancing out the changes within a tiny range:

• \( f(1.5) = 0.18 \)
• \( f(2.5) = 0.40 \)

The approximate instantaneous rate of change then came out to be:

• \( \frac{0.40 - 0.18}{2.5 - 1.5} = 0.22 \)

Meaning the function is increasing at a rate of 0.22 units at exactly \( x=2 \). This approach is similar to zooming in on a curve so closely that it appears to be a straight line.

Logarithms are the inverse operations of exponentiation. They answer the question: "To what power must we raise the base to get the number?" In the case of base 10 logarithms, we are dealing with the power to which the number 10 must be raised to get a particular number.
For example:

• \( \log_{10}(1) = 0 \) because \( 10^0 = 1 \)
• \( \log_{10}(10) = 1 \) because \( 10^1 = 10 \)
• \( \log_{10}(100) = 2 \) because \( 10^2 = 100 \)

In the problem you're studying, \( f(x) = \log_{10}(x) \), this defines the function whose values provide insights into relationships involving powers of 10. The logarithm base 10 function is widely used in science, engineering, and mathematics to simplify calculations involving exponential relationships. Understanding how this function behaves, such as rates of change, can offer powerful insights into various phenomena.