Consider the following non-linear equation where x is the independent variable, y is the dependent variable, and A and B are constants.[math: y=−AxB] Which of the following is the resulting constant (vertical axis-intercept) in linearised form?

Question

BlackTom AI · Accepted Answer

Treat the constant prefactor as (-A) and take base-10 logarithms to linearise. Starting from y = −A x^B, write the product inside a single log and use log laws:

1) Knowledge points:
- Logarithm rules: log10(uv) = log10(u) + log10(v) and log10(x^B) = B·log10(x).
- Linearised power law: y = C x^B becomes log10(y) = log10(C) + B·log10(x). The intercept on a plot of log10(y) vs log10(x) is log10(C).

2) Step-by-step reasoning and calculations:
y = −A x^B
Take base-10 logs of both sides (assuming the argument of the log is positive, i.e. the prefactor used is positive):
log10(y) = log10(−A x^B)
Apply product rule: log10(y) = log10(−A) + log10(x^B)
Apply power rule: log10(y) = log10(−A) + B·log10(x)

So in the linear form log10(y) = (intercept) + (slope)·log10(x), the intercept is log10(−A) and the slope is B.

3) Intermediate steps shown above for clarity:
- log10(−A x^B) = log10(−A) + log10(x^B)
- log10(x^B) = B·log10(x)

4) Why the correct answer is correct:
The intercept of the log-log plot is the logarithm of the multiplicative constant in the original power law. The multiplicative constant here is (−A), so the intercept is log10(−A).

5) Why other options are incorrect:
- a. log10(B): B is the exponent (slope on the log-log plot), not the intercept.
- b. exp(B): Exponentiating B is unrelated to the intercept in log10-linearisation.
- c. B: Again, B is the slope, not the intercept.
- e. −A: −A is the prefactor in the original equation, but after linearising with logs the intercept is its logarithm, not the raw value.

Note on sign/domain: taking log10(y) requires y>0, so the algebra assumes the prefactor used inside the log (−A) and y are positive (i.e., appropriate sign conventions). If one instead plots log10(−y) vs log10(x) when the measured y is negative, the intercept would be log10(A).

Confirming: the vertical-axis intercept in linearised form is log10(−A), matching the provided answer.

Consider the following non-linear equation where x is the independent variable, y is the dependent variable, and A and B are constants.[math: y=−AxB] Which of the following is the resulting constant (vertical axis-intercept) in linearised form?Single choice

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