Checking domains and the reasonableness of model predictions | Year 11

01 / LEARN

Understand the idea.

A valid manipulation is not the final check

Start with the restrictions of the original question: logarithm arguments must be positive, denominators must be nonzero, and exponential bases must be valid. Algebra can produce candidate solutions, but it cannot make a forbidden input permissible.

Combining logarithms can enlarge the apparent domain of an expression. Record the original restrictions before simplifying, then test each candidate in the original equation. This distinguishes a valid solution from an artefact of the transformation.

A model needs a useful contextual domain

A formula can be mathematically defined for every real time yet describe an experiment only over a limited interval. State what the input represents, its unit, and which times the evidence supports before interpreting a prediction.

Check practical constraints as well as arithmetic: counts require sensible whole-number reporting, money needs suitable precision, and a finite container has a capacity. A forecast outside the observed range needs justification that the model assumptions still hold.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Check a log-equation candidate

Solve $log (x + 5) - log (x - 1) = log 3$ .

Worked solution

1
Require both original log arguments to be positive.
$x + 5 > 0, x - 1 > 0 ⟹ x > 1$
Both conditions must hold simultaneously before applying the quotient law.
2
Combine the logarithms.
$log (\frac{x + 5}{x - 1}) = log 3 ⟹ \frac{x + 5}{x - 1} = 3$
Equal logarithms of positive arguments have equal arguments.
3
Solve the resulting linear equation.
$x + 5 = 3 (x - 1) ⟹ 8 = 2 x ⟹ x = 4$
The original domain ensures $x - 1 \neq 0$ , so multiplication by it is allowed.
4
Verify the candidate in the original equation.
$log (4 + 5) - log (4 - 1) = log 9 - log 3 = log 3$
The candidate satisfies $x > 1$ and reproduces the required right-hand side.

Answer

x=4

EXAMPLE 02

Separate a formula from its use

A culture model N(t)=200 times 1.4 to the power t is fitted during the first 6 hours. Its container holds at most 2000 cells. Evaluate a prediction after 10 hours.

Worked solution

1
Calculate the formula output at ten hours.
$N (10) = 200 {(1.4)}^{10} \approx 5785.09$
Substitution gives the numerical prediction made by the fitted exponential rule.
2
Compare it with the physical capacity.
$5785.09 > 2000$
The predicted number exceeds the maximum the stated container can hold.
3
Compare the time with the fitted interval.
$10 > 6$
Ten hours is beyond the first six hours used to fit the model, so this is extrapolation.
4
Interpret the prediction.
$N (10) \approx 5785 cells by the formula$
The arithmetic is valid, but the fixed-growth assumption cannot give a credible ten-hour forecast under the stated capacity limit.

Answer

About 5785 by the formula, but the forecast is not physically reasonable.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the real domain of $log (x^{2} - 9)$ .

Show answer & reasoning

$x < - 3$ or $x > 3$ .

Require $x^{2} - 9 > 0$ , or $(x - 3) (x + 3) > 0$ . The factors have the same sign only when $x < - 3$ or $x > 3$ ; the endpoints make the argument zero and are excluded.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the domain of $log (5 - 2 x)$ .
2 marks
Find the domain of $log x + log (x - 4)$ .
2 marks
Solve $log (x + 2) = log (3 x - 4)$ and check its domain.
3 marks
A model is $V (t) = 900 {(0.8)}^{t}$ for an asset bought at $t = 0$ . Explain why $t = - 2$ is not a prediction of its resale value after purchase.
2 marks
A population model gives 47.826 individuals. How should a reported prediction be written, and why?
2 marks
A student solves $log x + log (x - 3) = 1$ and gives $x = - 2$ because the combined product is 10. Explain precisely why this is invalid.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the domain of $log (5 - 2 x)$ .
2 marks
Require $5 - 2 x > 0$ .
$x < 2.5$
Find the domain of $log x + log (x - 4)$ .
2 marks
Both arguments must be positive.
$x > 4$
Solve $log (x + 2) = log (3 x - 4)$ and check its domain.
3 marks
Domain $x > \frac{4}{3}$ ; equate arguments: $x + 2 = 3 x - 4$ , so $x = 3$ .
At $x = 3$ both arguments equal 5.
$x = 3$
A model is $V (t) = 900 {(0.8)}^{t}$ for an asset bought at $t = 0$ . Explain why $t = - 2$ is not a prediction of its resale value after purchase.
2 marks
The formula is mathematically defined at -2, but the stated post-purchase context uses $t$ at least zero.
It refers to two years before purchase.
A population model gives 47.826 individuals. How should a reported prediction be written, and why?
2 marks
Individuals are counted in whole numbers. The unrounded value is useful in calculations but not as a literal count.
About 48 individuals, with the model assumptions stated.
A student solves $log x + log (x - 3) = 1$ and gives $x = - 2$ because the combined product is 10. Explain precisely why this is invalid.
3 marks
Although their product is positive, neither original logarithm is real. Original domain $x > 3$ excludes -2.
The original arguments -2 and -5 are not positive.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

In this lesson

Understand the idea.

A valid manipulation is not the final check

A model needs a useful contextual domain

See the method in action.

Check a log-equation candidate

Require both original log arguments to be positive.

Combine the logarithms.

Solve the resulting linear equation.

Verify the candidate in the original equation.

Separate a formula from its use

Calculate the formula output at ten hours.

Compare it with the physical capacity.

Compare the time with the fitted interval.

Interpret the prediction.

Try it yourself.

Check your understanding

Keep practising