A logarithm answers an exponent question. If b^y=x, then log_b(x)=y. This makes logarithmic and exponential functions inverse representations of the same relationship.
1. Read logarithms as exponent statements
log₂8=3 because 2³=8. log₁₀0.01=−2 because 10^−2=0.01. ln(e^5)=5 because ln uses base e.
2. Domain comes first
For real logarithms, the argument must be positive. Thus log(x−4) requires x>4. The logarithm of zero or a negative real number is not defined in this real-valued setting.
3. Product law
For positive M and N, log_b(MN)=log_bM+log_bN. The law follows because multiplying powers of the same base adds exponents.
4. Quotient law
log_b(M/N)=log_bM−log_bN for positive M and N. Division subtracts exponents.
5. Power law
log_b(M^k)=k log_bM when the real expressions are defined. The exponent becomes a multiplier.
6. There is no corresponding sum law
log(M+N) is not generally logM+logN. For example, log₁₀(10+10)=log₁₀20, whereas log₁₀10+log₁₀10=2. These are different.
7. Graph as an inverse
y=b^x and y=log_bx are reflections of each other in y=x. The exponential point (0,1) corresponds to logarithmic point (1,0).
The exponential horizontal asymptote y=0 becomes the logarithmic vertical asymptote x=0.
8. Base controls direction
When b>1, y=log_bx increases. When 0<b<1, it decreases. In both cases the domain is x>0 and the range is all real numbers.
9. Worked simplification
Simplify 2lnx+lny−lnz for positive x,y,z.
2lnx=ln(x²). Therefore the expression is ln(x²y/z).
10. Common mistakes
- Applying a false log sum law.
- Ignoring the positive-argument condition.
- Confusing ln with log base 10.
- Forgetting that an inverse graph swaps coordinates.
11. Practice
- Evaluate log₂32.
- Evaluate log₁₀0.001.
- Expand ln(3x²/y) for positive x,y.
- Condense 3lna−2lnb.
- State the domain of log(x+7).
12. Answers
- 5.
- −3.
- ln3+2lnx−lny.
- ln(a³/b²), for positive a,b.
- x>−7.
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