For AI agents: use /llms.txt for the Nakafa content index.
Logarithms undo exponentiation. If , then . Every example below uses a positive logarithm argument and a base that satisfies and .
Use positive arguments for every rule below. For the power rule, the base of the power remains positive even if the power itself could also be evaluated at a negative number.
The logarithm of a product equals the sum of the logarithms of each number.
Example:
The logarithm of a quotient equals the difference between the logarithm of the numerator and the logarithm of the denominator.
Example:
The logarithm of a number raised to a power equals the power multiplied by the logarithm of that number.
Example:
Check the base conditions before applying a rule: both the original and replacement bases must be positive and different from one.
For example, changing from base two to base ten preserves the value. The denominator is nonzero because the original base is not one.
Use this identity to rewrite a logarithm in a base that is available or easier to evaluate.
Example:
Because , the numerator is exactly twice the denominator:
If , then
Two numbers that have the same logarithmic value (with the same base) must be the same number. For example, and both sides equal , so the arguments and are equal. The rule works because each value of a logarithm with a fixed base comes from exactly one argument.
The comparison depends on the direction of the graph. A base above one preserves the ordering of the arguments. A base between zero and one reverses it.
Here an identity meets an algebraic expression, so the expression has to reach a suitable form first. Work step by step and note which identity you use at each step.
The identities earn their place when an expression must be simplified before it can be solved.
Use the product and then quotient identities to combine the three terms.
Simplify
Solution:
The initial condition is a positive argument, so the input must be above zero. The solution nine satisfies that condition.
Find the value of if
Solution:
In a model the unknown quantity sits in the exponent, and a logarithm brings it down. The examples below show which identity each one needs.
Logarithmic identities also appear when a model contains exponential growth or decay.
This model compares an amplitude ratio, using logarithms to base ten.
A simplified magnitude scale can compare a measured seismic-wave amplitude with a reference amplitude :
Where:
Example: Suppose a measured amplitude is times the reference amplitude. What magnitude does the simplified model give?
Solution:
The amplitude ratio therefore corresponds to magnitude in this simplified model. Actual earthquake-magnitude calculations also account for the instrument, wave type, and source distance.
In an idealized first-order charging model, the time needed to reach a fraction of the maximum capacity is:
Where:
Example: Determine the time required to charge a battery from empty to full. Assume .
Solution:
Within this idealized model, reaching takes approximately . Real charging systems can change their rate near full capacity.
The charging model domain runs from zero stored charge to less than the maximum.
In this model, full charge is approached without being reached in finite time. Comparing targets shows why an extra percent near full charge can take longer.
| Charge target | Time (minutes) |
|---|---|
The fixed model rate is an assumption over the modeled interval. The car value halves every five years.
If the instantaneous depreciation rate is proportional to a car's current value, an exponential model has the form:
In this model, is the car's value at time .
Example: A new car costs and its modeled value after is . Assuming the same exponential depreciation rate continues, determine the modeled value after .
Solution:
Measure in millions of rupiah and in years. At purchase:
From these results, the car's price at any time is:
The car's price after of use is:
The three problems each need a different rewrite, so write down which rule you need before you calculate. Then compare your route with the route the solution takes.
Problem 1
Simplify:
Problem 2
If and , find the value of
Problem 3
Find the value of if
Answer 1
Each number is a power of . Evaluate the three logarithms before combining them:
The simplified value is .
Answer 2
Use the product identity because the target contains :
.
Answer 3
Use the quotient identity on the right-hand side, then use equality of logarithms with the same base:
The logarithm requires . Since , the solution is valid.
Published: . Updated: .
