Logarithm laws, negative powers and equation domains
| English | Français |
|---|---|
| argument/ˈɑːɡjuːmənt/ | argument |
Combining two logs can produce a polynomial with an extra root. Which original input condition rejects it?
- Combining two logs can produce a polynomial with an extra root. Which original input condition rejects it?
- This lesson studies argument 真数: The input inside a logarithm, which must be positive for a real logarithm.
Choose the mathematical structure
- For a>0, a≠1 and positive u,v: log_a(uv)=log_a u+log_a v, log_a(u/v)=log_a u−log_a v, and log_a(u^k)=k log_a u for real k. These follow from exponent multiplication/division/powers. In particular k=−1 gives a reciprocal and k=−1/2 gives an inverse square root. Solve a^x=b, with b>0, by x=ln b/ln a.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines argument?
The input inside a logarithm, which must be positive for a real logarithm.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For x>0, ln(x^(-1))=−ln x and ln(x^(-1/2))=−(1/2)ln x. Solve ln(x−1)+ln(x+1)=ln8: the original domain is x>1. Combining gives ln(x²−1)=ln8, hence x²=9 and candidates ±3; only x=3 satisfies the original arguments. For 3^x=20, x=ln20/ln3. If 2^(2x)−2^x−2=0, let u=2^x>0: (u−2)(u+1)=0, so u=2 gives x=1, while u=−1 is impossible. Also ln(2+3)=ln5, whereas ln2+ln3=ln6; a sum inside a log cannot be split this way.
Logarithm laws, negative powers and equation domains
For a>0, a≠1 and positive u,v: log_a(uv)=log_a u+log_a v, log_a(u/v)=log_a u−log_a v, and log_a(u^k)=k log_a u for real k
State the valid input, plotted variable or time unit before using the logarithmic/exponential relationship.
Solve 3^x=27.
27=3³, so x=3.
Test a tempting shortcut
- Combining or expanding logs does not erase their original domains. ln(u²)=2ln|u| for u≠0; writing 2lnu additionally needs u>0. Do not take a real logarithm of a zero or negative right-hand side in a^x=b. For base 1, 1^x=b has all real x when b=1 and no solutions otherwise; the log quotient is not available.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every polynomial root obtained after combining logarithms satisfies the original log equation. This claim is false. Explain which definition or assumption it violates.
For x>0 write ln(x^(-1/2))=klnx. Find k.
The power law takes exponent −1/2 outside: k=−1/2.
Every polynomial root obtained after combining logarithms satisfies the original log equation.
Combining or expanding logs does not erase their original domains. ln(u²)=2ln|u| for u≠0; writing 2lnu additionally needs u>0. Do not take a real logarithm of a zero or negative right-hand side in a^x=b. For base 1, 1^x=b has all real x when b=1 and no solutions otherwise; the log quotient is not available.
Interpret a new situation
- List positivity restrictions first, use one common base, then apply a valid law. After algebra, substitute every candidate into the original logarithmic/exponential equation. Keep log quotients exact unless a decimal is requested. Changing the log base does not change an exponential equation’s solution when both quotient logs use the same valid base.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the valid x for ln(x−1)+ln(x+1)=ln8.
x²=9 gives ±3, but both original log arguments are positive only for x>1; accept 3.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · F. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The input inside a logarithm, which must be positive for a real logarithm. Choose the relationship, show the method, check its assumptions and interpret the result.