Simplified Quantitative Formulas: Logarithms, Surds, and Indices
- Indices: X^m × X^n = X^(m+n); X^m / X^n = X^(m–n); (X^m)^n = X^(mn).
- Negative/Fractional: X^–m = 1/X^m; X^(1/a) = a-th root of X.
- Surds: Like surds can be added/subtracted. Conjugate: (a+√b)(a–√b) = a²–b.
- Rationalizing: Multiply by conjugate to remove roots from denominator.
- Logarithms: log_a N = x means a^x = N. log_a 1 = 0; log_a a = 1.
- Product Rule: log_a(XY) = log_a X + log_a Y.
- Quotient Rule: log_a(X/Y) = log_a X – log_a Y.
- Power Rule: log_a(X^k) = k × log_a X.
- Change of Base: log_b M = log_a M / log_a b.
What do these mean? (Super Simple Explanations & Examples)
- Indices:
- Xm means X multiplied by itself m times.
- Example: 23 = 2 × 2 × 2 = 8.
- Negative Index: 2-2 = 1/(22) = 1/4.
- Fractional Index: 91/2 = √9 = 3. - Surds:
- A surd is an irrational root (like √2, √3) that cannot be simplified to a rational number.
- Example: √2 ≈ 1.414, which cannot be written as a simple fraction.
- Like Surds: 2√3 and 5√3 can be added (result: 7√3), but 2√2 and 3√3 cannot be combined. - Logarithms:
- A logarithm answers: "To what exponent must the base be raised, to get this number?"
- Example: log2(8) = 3, because 23 = 8.
- Another Example: log10(100) = 2, because 102 = 100.
- loga(1) = 0 for any base a, because a0 = 1.
- loga(a) = 1 for any base a, because a1 = a.
Introduction to Logarithms
A logarithm is the inverse operation of exponentiation. If b^x = y, then log_b(y) = x.
Basic Definition
If b^x = y, then log_b(y) = x
where:
- b is the base (b > 0, b ≠ 1)
- y is the argument (y > 0)
- x is the exponent
Q1
Evaluate log₂(8)
Solution:
We need to find x such that 2^x = 8
2^3 = 8
Therefore, log₂(8) = 3
Q2
Evaluate log₅(125)
Solution:
We need to find x such that 5^x = 125
5^3 = 125
Therefore, log₅(125) = 3
Properties of Logarithms
Logarithms have several fundamental properties that are essential for solving problems and simplifying expressions.
1. Basic Properties
- log_b(1) = 0 (Any base raised to 0 is 1)
- log_b(b) = 1 (Base raised to 1 is itself)
- log_b(b^x) = x (Logarithm and exponent cancel each other)
- b^(log_b(x)) = x (Exponent and logarithm cancel each other)
Example 1
Evaluate: log₅(1) + log₅(5)
Solution:
Using basic properties:
log₅(1) = 0 (since 5⁰ = 1)
log₅(5) = 1 (since 5¹ = 5)
Therefore, log₅(1) + log₅(5) = 0 + 1 = 1
2. Product Rule
log_b(xy) = log_b(x) + log_b(y)
The logarithm of a product is the sum of the logarithms.
Example 2
Simplify: log₂(8) + log₂(4)
Solution:
Using product rule:
log₂(8) + log₂(4) = log₂(8 × 4)
= log₂(32)
= log₂(2⁵)
= 5
3. Quotient Rule
log_b(x/y) = log_b(x) - log_b(y)
The logarithm of a quotient is the difference of the logarithms.
Example 3
Simplify: log₃(27) - log₃(3)
Solution:
Using quotient rule:
log₃(27) - log₃(3) = log₃(27/3)
= log₃(9)
= log₃(3²)
= 2
4. Power Rule
log_b(x^n) = n·log_b(x)
The logarithm of a power is the exponent times the logarithm of the base.
Example 4
Simplify: log₂(8⁴)
Solution:
Using power rule:
log₂(8⁴) = 4·log₂(8)
= 4·log₂(2³)
= 4·3
= 12
5. Root Rule
log_b(√x) = (1/n)·log_b(x)
The logarithm of a root is the logarithm of the number divided by the root index.
Example 5
Simplify: log₃(√27)
Solution:
Using root rule:
log₃(√27) = (1/2)·log₃(27)
= (1/2)·log₃(3³)
= (1/2)·3
= 1.5
6. Change of Base Rule
log_b(x) = log_a(x)/log_a(b)
Allows conversion between different bases.
Example 6
Evaluate: log₅(25) using base 2
Solution:
Using change of base rule:
log₅(25) = log₂(25)/log₂(5)
= 4.64/2.32
= 2
7. Reciprocal Rule
log_b(1/x) = -log_b(x)
The logarithm of a reciprocal is the negative of the logarithm.
Example 7
Simplify: log₂(1/8)
Solution:
Using reciprocal rule:
log₂(1/8) = -log₂(8)
= -log₂(2³)
= -3
8. Base Change Identity
log_b(x) = 1/log_x(b)
Relates logarithms with swapped base and argument.
Example 8
If log₂(8) = 3, find log₈(2)
Solution:
Using base change identity:
log₈(2) = 1/log₂(8)
= 1/3
9. Combined Properties
These properties can be combined to solve complex problems.
Example 9
Simplify: log₂(16) + log₂(4) - log₂(8)
Solution:
Using product and quotient rules:
log₂(16) + log₂(4) - log₂(8) = log₂((16 × 4)/8)
= log₂(64/8)
= log₂(8)
= log₂(2³)
= 3
Change of Base Formula
The change of base formula allows us to convert logarithms from one base to another.
Change of Base Formula
log_b(x) = log_a(x)/log_a(b)
where:
- a, b are positive numbers not equal to 1
- x is a positive number
Q1
Evaluate log₅(25) using base 2
Solution:
Using change of base formula:
log₅(25) = log₂(25)/log₂(5)
= 4.64/2.32
= 2
Natural Logarithms
Natural logarithms use e (approximately 2.71828) as their base.
Natural Logarithm Properties
- ln(1) = 0
- ln(e) = 1
- ln(e^x) = x
- e^(ln x) = x
- ln(xy) = ln(x) + ln(y)
- ln(x/y) = ln(x) - ln(y)
- ln(x^n) = n·ln(x)
Q1
Evaluate ln(e³)
Solution:
Using the property ln(e^x) = x
ln(e³) = 3
Applications of Logarithms
Logarithms have numerous applications in various fields.
Common Applications
- pH Scale: pH = -log₁₀[H⁺]
- Richter Scale: M = log₁₀(A/A₀)
- Decibel Scale: dB = 10·log₁₀(I/I₀)
- Compound Interest: A = P(1 + r/n)^(nt)
Q1
Calculate the pH of a solution with [H⁺] = 1 × 10⁻⁷ M
Solution:
pH = -log₁₀[H⁺]
pH = -log₁₀(1 × 10⁻⁷)
pH = -(-7)
pH = 7
Advanced Concepts
More complex logarithm concepts and their applications.
Advanced Properties
- log_b(x) = 1/log_x(b)
- log_b(x) = ln(x)/ln(b)
- log_b(x) = log₁₀(x)/log₁₀(b)
- log_b(x) = -log_b(1/x)
Q1
Solve for x: log₂(x) + log₂(x-2) = 3
Solution:
log₂(x) + log₂(x-2) = 3
log₂(x(x-2)) = 3
x(x-2) = 2³
x² - 2x = 8
x² - 2x - 8 = 0
(x-4)(x+2) = 0
x = 4 or x = -2
Since x must be positive, x = 4
Q2
Solve for x: 2^(x+1) = 5^(x-1)
Solution:
Take log of both sides:
log(2^(x+1)) = log(5^(x-1))
(x+1)log(2) = (x-1)log(5)
x·log(2) + log(2) = x·log(5) - log(5)
x(log(2) - log(5)) = -log(5) - log(2)
x = (-log(5) - log(2))/(log(2) - log(5))
x ≈ 2.32
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