Semi-log Plots
Introduced| English | Meaning |
|---|---|
| exponential/ˌekspəˈnenʃl/ | Describing a relationship in which a variable occurs in an exponent. |
| semi-log plot/ˈsemi lɒɡ plɒt/ | A graph with one logarithmic axis and one linear axis. |
| logarithmic scale/ˌlɒɡəˈrɪθmɪk skeɪl/ | logarithmic scale |
| slope/sləʊp/ | The change in a line's vertical coordinate divided by the corresponding change in its horizontal coordinate. |
| intercept/ˌɪntəˈsept/ | The coordinate at which a graph crosses a specified coordinate axis. |
Follow a repeated growth factor
- A culture has counts $4,8,16,32,64$ at times $0,1,2,3,4$ hours. Equal time steps multiply the count by 2, so an exponential 指数 model is $y=4\cdot2^x$.
- Ordinary axes show a rising curve. A semi-log plot 半对数图 can show the same values as a straight line. Changing the scale changes their positions on the page, not the counts or their units. Between observations, this model assumes continuous growth; predicted counts can be approximate.
Read a logarithmic axis
- A logarithmic scale 对数刻度 gives equal space to equal factors: $1,10,100,1000$ are equally spaced. Halfway between 10 and 100 is $\sqrt{10\cdot100}\approx31.623$, not 55.
- A semi-log plot has exactly one logarithmic axis. For the exponential examples here, the horizontal axis is linear and the vertical axis is logarithmic. A standard log axis cannot show zero or negative values.
A semi-log plot uses a logarithmic scale on…
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Exactly one axis is logarithmic. Its orientation matters when interpreting a straight pattern.
What output lies halfway between 10 and 100 on a logarithmic scale? Give three decimal places.
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The halfway value is the geometric mean √(10×100)=√1000≈31.623.
Explain why an exponential straightens
- For $y=ab^x$ with $a>0$ and $b>0$, take a logarithm of base $q>1$. Then $\log_q y=\log_q a+x\log_q b$, which is linear in $x$.
- On a log-scaled vertical axis, ticks can still show the original $y$ values. On an ordinary axis labelled $\log_q y$, the plotted coordinates are the transformed values. Both give the same straight-line pattern when their scales match; do not take a log twice.
Which original output can be placed on a standard logarithmic vertical axis?
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A real logarithm needs a positive argument. Zero and negative outputs have no finite real log coordinate.
Recover the original coefficients
- In transformed coordinates $z=\log_q y=mx+c$, the slope 斜率 is $m=\log_q b$ and the intercept 截距 is $c=\log_q a$. Recover $a=q^c$ and $b=q^m$.
- The original initial value is $a$, not $c$. For $y=4\cdot2^x$ with base 2, the transformed line is $z=x+2$: slope 1, intercept 2, initial count 4. The slope describes a log change per input unit, not an ordinary increase in $y$.
Compare ordinary and logarithmic vertical scales
Compare the same values for growth, decay, a constant, a linear contrast and an offset exponential. Read original outputs and transformed coordinates.
For $y=ab^x$, with $a>0,b>0$, the ____ of $z=\log_q y$ against $x$ equals $\log_q b$.
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The slope is the log change per input unit. The factor in the original output is $b=q^m$, with the stated log base and units.
A base-10 transformed line has intercept $c=2$. What is the original output at $x=0$?
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Exponentiate the intercept: a=10²=100.
Build a model from two observations
- Suppose positive observations are $(0,100)$ and $(4,10000)$. On base-10 transformed axes they are $(0,2)$ and $(4,4)$, giving $m=(4-2)/4=0.5$ and $c=2$.
- Therefore $\log_{10}y=0.5x+2$, so $y=100\cdot10^{0.5x}=100(\sqrt{10})^x$. The output multiplies by 100 over four input units, and by $\sqrt{10}$ per unit. At $x=2$, the model predicts 1000.
Positive observations are $(0,100)$ and $(4,10000)$. What is the slope of $\log_{10}y$ against $x$?
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The transformed outputs are 2 and 4, so slope=(4−2)/(4−0)=0.5.
Which model corresponds to $\log_{10}y=0.5 x+2$?
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Exponentiate the whole right side: y=10²×10^(0.5 x).
Interpret growth, decay and a constant
- With $q>1$, positive transformed slope means growth and negative slope means decay. If $\log_{10}y=2-0.25t$, then $y=100(10^{-0.25})^t$: the factor per time unit is about 0.5623, a decrease of about $43.77\%$.
- A zero slope gives a positive constant, such as $y=4$. It is the case $b=1$, so a horizontal straight line alone does not show growth or decay. A nonconstant ordinary linear function generally curves when its output axis is log-scaled.
Every ordinary linear function, including nonconstant ones, stays straight when its positive output is plotted on a logarithmic vertical axis.
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False. Nonconstant ordinary linear functions generally curve on log-output axes. A positive constant remains horizontal; positive exponentials form straight patterns.
For $\log_{10}y=2-0.25 t$, give the output factor per time unit to four decimal places.
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The factor is 10^(−0.25)≈0.5623; −0.25 is the transformed slope, not a 25% decrease.
Keep the base and time units explicit
- Changing the log base changes the numerical slope and intercept, but not the original model. For $4\cdot2^x$, base 10 gives $m=\log_{10}2\approx0.3010$ and $c=\log_{10}4\approx0.6021$; base 2 gives $m=1,c=2$.
- If $x$ is hours and $u$ is minutes, then $x=u/60$. The base-2 line becomes $z=u/60+2$, with slope $1/60$ per minute. A slope comparison is meaningful only with the same log base and input units.
Select all true statements for positive exponential data on a logarithmic vertical axis and a linear horizontal axis.
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Exactly one axis is logarithmic; an ideal positive exponential gives a line. Its slope depends on the log base and input units. Other data need not straighten.
The line is $\log_2 y=x+2$ when $x$ is hours. What is its slope when time is in minutes?
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One minute is 1/60 hour; the log output changes by 1/60 per minute.
Use a line to find a time
- For the decay model $\log_{10}y=2-0.25t$, half the initial amount is $y=50$. Solve $\log_{10}50=2-0.25t$, giving $t=\dfrac{\log_{10}2}{0.25}\approx1.204$ time units.
- Recover a value with an exponential and a time with the line equation. Keep full coefficients until the final rounding, and distinguish interpolation within the observed time range from extrapolation beyond it.
For $\log_{10}y=2-0.25 t$, when does the amount first reach 50? Give three decimal places.
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Half the initial 100 means t=log₁₀2/0.25≈1.204 time units.
Check which axis is logarithmic
- If the horizontal axis is logarithmic and the vertical axis is linear, a model such as $y=2\log_{10}x+1$ forms a straight line against $\log_{10}x$. The input must be positive; this orientation points to a logarithmic model.
- Thus “straight on a semi-log plot” needs an axis description. Logarithmic vertical axes test positive exponential output patterns; logarithmic horizontal axes test logarithmic input patterns. A plot with both axes logarithmic is a different kind of plot.
Data form a line when plotted as ordinary $y$ against $\log_{10}x$. Which is a matching form?
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The logarithm is applied to the input, so a logarithmic model is linear in log(input).
Check evidence beyond a straight appearance
- Two positive observations always determine a line after their outputs are logged, if their inputs differ. They do not establish that an exponential model is reliable. With more data, inspect deviations, context and the observed range; noisy exponential data need not be perfectly straight.
- Fitting a line to $\log y$ minimises squared errors in log output, not squared errors in the original $y$. It is not automatically a more accurate fit for the question being asked. Check predictions in the original units too.
Several positive observations lie close to a nonhorizontal line with a linear horizontal axis and a logarithmic vertical axis. Which model family is a candidate to check?
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An exponential model is a candidate. Check deviations and context before judging reliability; a near-line pattern is evidence, not proof.
Two positive observations at different inputs are enough to prove that exponential extrapolation is reliable.
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Two transformed points always determine a line. Other observations and the context are needed to assess a model.
Handle offsets and invalid values honestly
- A vertical offset changes the test: $y=4\cdot2^x+4$ generally does not straighten when $y$ itself is logged. If a baseline of 4 is justified, plotting $\log(y-4)$ can reveal the exponential part, where $y-4>0$.
- Do not remove nonpositive observations or replace them by positive values just to make a log plot work. First check the measurements and whether a justified model or transformation fits the context. A scale change cannot validate a model or make extrapolation safe.
Read the axis labels and base first. For positive outputs on a logarithmic vertical axis, $\log_q y=mx+c$ corresponds to $y=q^c(q^m)^x$. Interpret the factor per input unit, check other observations and retain the original units in predictions.
Explain the same value on two scales
- For $y=4\cdot2^x$, find the output at $x=3$. Then find its vertical coordinate if the axis is labelled $z=\log_2 y$.
- Explain why those two numbers differ and how the original initial value and growth factor are recovered from $z=x+2$. State which axis is logarithmic when you describe the straight pattern.
For $y=4\cdot 2^x$, find $y$ at $x=3$ and the corresponding coordinate $z=\log_2 y$. Explain the straight-line equation and recover the original initial value and growth factor.
Compute the original value, then its logarithm; recover coefficients by exponentiating in the same base.
For $y=4\cdot 2^x$, what is the original output $y$ at $x=3$?
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The original value is 4×2³=32. Its base-2 logarithm is 5; the scale does not change the output.