Semi-log Plots

Worked example

1. 01

   (a) 30% of the way means $\log y = 2 + 0.3 = 2.3$.

2. 02

   Undo the log: $y = 10^{2.3} \approx 199.5$. The point is at about 200, not at $100 + 0.3 \times 900 = 370$.

3. 03

   (b) $\log 400 \approx 2.602$, so 400 sits about 60% of the way up the band. (200 sits at 30.1% and 500 at 69.9%; the gridlines are NOT evenly spaced.)

Worked example

1. 01

   Check linearity: the differences of $\log y$ are all between about 0.16 and 0.20, so ($t$, $\log y$) is close to a line. The data is close to exponential.

2. 02

   Pick two points far apart, $t = 1 (y = 390)$ and $t = 7 (y = 4450)$, and find the slope of the line through them:

   $$m = \frac{\log 4450 - \log 390}{7 - 1} \approx 0.1762, \qquad b = 10^{m} \approx 1.5004$$
3. 03

   Intercept: $\log a = \log 390 - 0.1762 \cdot 1 \approx 2.4148$, so $a = 10^{2.4148} \approx 259.9$. Estimated model: $y \approx 259.9(1.5004)^{t}$, which predicts about 6,677 at $t = 8$ (actual: 6,490).

Worked example

1. 01

   Fit a LINEAR regression to the points ($t$, $\log y$): $\log y \approx 2.4046 + 0.1767t$. (TI-84: put $\log y$ in L3 and use LinReg on L1, L3. Desmos: add a column $\log(y_{1})$ and type $\log(y_{1})$ ~ $m \cdot x_{1}$ + c.)

2. 02

   Undo the log. The slope becomes the base $b = 10^{0.1767}$ and the intercept becomes $a = 10^{2.4046}$:

   $$\log y \approx 2.4046 + 0.1767\,t \;\Rightarrow\; y \approx 10^{2.4046} \cdot \left(10^{0.1767}\right)^{t} \approx 253.844\,(1.5022)^{t}$$
3. 03

   Compare: the hand estimate $(259.9, 1.5004)$ and the regression $(253.844, 1.5022)$ are close. The regression is better because it uses every point, not just two.

   This is exactly the ExpReg model from 2.5B, and that is no coincidence: exponential regression works by fitting a line to the logarithms of the data. That is also where the $r \approx 0.99983$ in 2.5B came from: it measures how straight the ($t$, $\log y$) points are. (It's also what Desmos's “Log Mode” does.) With ln instead of log you get $\ln y = \ln a + t \ln b$ and the same $a$ and $b$.

Worked example

1. 01

   The slope of the line is the change in $\log y$ over the change in $x$:

   $$m = \frac{\log 8000 - \log 500}{6 - 2} = \frac{\log 16}{4} \approx 0.301 \;\Rightarrow\; b = 10^{0.301} = 2$$
2. 02

   Or directly: from $x = 2$ to $x = 6$, $y$ is multiplied by $8000/500 = 16$ in 4 steps, so $b^{4} = 16$ and $b = 2$. (The base of an exponential must be positive, so −2 is not possible, as in 2.5A.)

3. 03

   $a = 500 \div 2^{2} = 125$, so $y = 125 \cdot 2^{x}$. Check: $125 \cdot 2^{6} = 8000$ ✓.

Worked example

Worked example

Worked example

Worked example

Worked example

Worked example