To find the equation for the given graph on a semi-logarithmic scale, let's analyze the information.

Understanding the Semi-Log Scale

The $y$-axis is labeled as $\log(y)$, indicating that this is a semi-logarithmic plot, where:

• The $x$-axis is in a linear scale.
• The $y$-axis represents $\log(y)$, meaning each unit on the $y$-axis corresponds to an exponential increase in $y$.

This suggests the function is of the form: $y = a \cdot b^x$ where $\log(y) = \log(a) + x \cdot \log(b)$, and thus, the graph of $\log(y)$ versus $x$ should be a straight line.

Analyzing the Line

On the graph:

1. Identify two points on the line, as we need them to determine the slope and intercept.

   • Looking at the image, the line seems to pass through points approximately at $(0, -3)$ and $(2, -2)$.

2. Using these points, calculate the slope $m$: $m = \frac{-2 - (-3)}{2 - 0} = \frac{1}{2}$

3. The equation of the line in terms of $\log(y)$ is: $\log(y) = \frac{1}{2}x - 3$

Converting to Exponential Form

Now, rewrite this in terms of $y$:

1. Recall that $\log(y) = \frac{1}{2}x - 3$.
2. Converting from logarithmic to exponential form: $y = 10^{\frac{1}{2}x - 3}$
3. Simplify: $y = 10^{-3} \cdot 10^{\frac{1}{2}x} = \left(10^{-3}\right) \cdot \left(10^{\frac{1}{2}}\right)^x$
4. Thus, $a = 10^{-3}$ and $b = 10^{\frac{1}{2}}$.

So, the final equation is: $y(x) = 0.001 \cdot (10^{0.5})^x$

Let me know if you'd like more details or if you have questions about any part of the solution!

Here are 5 related questions to expand your understanding:

1. How does a semi-log scale differ from a log-log scale?
2. What types of real-world data are best represented by exponential functions?
3. How would the slope change if the function’s growth rate increased?
4. Why is the y-axis in log scale rather than linear scale for this graph?
5. How do we interpret the intercept in an exponential function on a semi-log scale?

Tip: When working with semi-log scales, keep in mind that the line's slope represents the growth rate of the exponential function.