Learning goal and prerequisites
Translate straight-line coordinates in log space into an exponential rule. Before starting, review exponential fit.
Method and conditions
Name the logarithm base and which axis is transformed. A graph of L=log₁₀(y) against t with L=mt+c means y=10^c(10^m)^t. Vertical differences are log ratios, not additive changes in y.
Worked example: show the decisions
If the plotted points are (0,1), (2,2) and (4,3), then m=1/2 and c=1. Therefore y=10×10^(t/2), giving original outputs 10,100,1000 at the labeled times. The factor of ten applies every two time units.
Contrast: why a plausible shortcut fails
On this display a vertical rise from 1 to 2 represents y changing from 10 to 100. Calling that an increase of one in y confuses plotted coordinates with the original measured quantity. A zero log coordinate represents y=1, not y=0.
Transfer: try before revealing the answer
A semi-log line is log₂(y)=6−2t. Find y(0), the one-unit multiplier and y(3).
Transfer solution and check
The initial output is 64, the multiplier is 2^(−2)=1/4, and y(3)=1. The decreasing line represents positive decay; it does not imply negative output. To compare log₁₀(A)=2t+1 with log₁₀(B)=t, subtract: log₁₀(A/B)=t+1. Their ratio reaches100 when t=1, not when either separate model reaches100.
A straight line on a semi-log display has vertical coordinate L=log₁₀(y) and equation L=1t+2. What factor multiplies y when t increases by 1?
- A one-unit t step adds 1 to the logarithm, so it multiplies the original output by 10^1.
- Using these quantities: 10^1 = 10.
- Check the original domain, requested quantity and units; do not substitute a related parameter.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Precalculus Course at a Glance