CELE Engineering Mathematics — Algebra and FundamentalsStudy Notes

Exponents and radicals are compact notation systems for repeated multiplication and fractional powers. Understanding these laws is essential because engineering formulas often involve powers (e.g., stress concentration factors, deflection coefficients, and flow rate calculations).

Fundamental Laws of Exponents:

1. Product Rule: $a^m \cdot a^n = a^{m+n}$

   • When multiplying powers with the same base, add the exponents.
   • Example: $x^3 \cdot x^5 = x^{3+5} = x^8$

2. Quotient Rule: $\frac{a^m}{a^n} = a^{m-n}$ (where $a \neq 0$)

   • When dividing powers with the same base, subtract the exponents.
   • Example: $\frac{y^7}{y^2} = y^{7-2} = y^5$

3. Power of a Power: $(a^m)^n = a^{mn}$

   • When raising a power to another power, multiply the exponents.
   • Example: $(2^3)^4 = 2^{3 \times 4} = 2^{12} = 4096$

4. Power of a Product: $(ab)^n = a^n b^n$

   • Distribute the exponent to each factor.
   • Example: $(3x)^2 = 3^2 x^2 = 9x^2$

5. Power of a Quotient: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ (where $b \neq 0$)

   • Distribute the exponent to both numerator and denominator.
   • Example: $\left(\frac{2}{5}\right)^3 = \frac{2^3}{5^3} = \frac{8}{125}$

6. Zero Exponent: $a^0 = 1$ (where $a \neq 0$)

   • Any non-zero number raised to the power of 0 equals 1.
   • Example: $7^0 = 1$; $(x^2)^0 = 1$

7. Negative Exponent: $a^{-n} = \frac{1}{a^n}$ (where $a \neq 0$)

   • A negative exponent means take the reciprocal.
   • Example: $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$; $x^{-3} = \frac{1}{x^3}$

Radicals and Fractional Exponents:

Radicals are written as $\sqrt[n]{a}$ (the $n$-th root of $a$) and are equivalent to fractional exponents: $\sqrt[n]{a} = a^{1/n}$.

• Square root: $\sqrt{a} = a^{1/2}$
• Cube root: $\sqrt[3]{a} = a^{1/3}$
• General fractional exponent: $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$

Radical Laws:

1. Product Rule: $\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$

   • Example: $\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}$

2. Quotient Rule: $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ (where $b \neq 0$)

   • Example: $\sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2}$

3. Power of a Radical: $(\sqrt[n]{a})^m = \sqrt[n]{a^m} = a^{m/n}$

   • Example: $(\sqrt{5})^4 = (5^{1/2})^4 = 5^2 = 25$

4. Rationalization: Remove radicals from denominators by multiplying by a convenient form of 1.

   • Example: $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Worked Example 1: Simplify $\frac{(2x^3)^2 \cdot (3x^{-1})^3}{(x^2)^4}$

Step 1: Apply the power of a product rule to each term:

• $(2x^3)^2 = 2^2 (x^3)^2 = 4x^6$
• $(3x^{-1})^3 = 3^3 (x^{-1})^3 = 27x^{-3}$
• $(x^2)^4 = x^8$

Step 2: Substitute into the original expression:

Step 3: Apply the quotient rule:

Worked Example 2: Simplify $\sqrt[3]{54} - \sqrt[3]{2}$

Step 1: Factor the radicand in the first term:

Step 2: Apply the product rule for radicals:

Step 3: Combine like radicals:

Logarithms are the inverse of exponential functions. They transform multiplicative operations into additive ones, making complex calculations manageable—a critical skill in engineering where compound growth, decay, and stress-strain relationships often follow exponential or logarithmic patterns.

Definition:

If $b^x = y$ (where $b > 0$, $b \neq 1$), then $\log_b y = x$.

• $b$ is the base
• $y$ is the argument
• $x$ is the logarithm (the exponent)

Common Bases:

• Base 10 (Common logarithm): $\log_{10} y$ or $\log y$ (used in engineering decibel calculations, pH scales)
• Base $e$ (Natural logarithm): $\log_e y$ or $\ln y$ (used in growth/decay, control systems)
• Base 2 (Binary logarithm): $\log_2 y$ (used in computer science and information theory)

Fundamental Logarithmic Laws:

1. Product Rule: $\log_b(MN) = \log_b M + \log_b N$

   • The logarithm of a product equals the sum of logarithms.
   • Example: $\log_2(8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5$
   • Verification: $8 \times 4 = 32 = 2^5$, so $\log_2 32 = 5$ ✓

2. Quotient Rule: $\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N$

   • The logarithm of a quotient equals the difference of logarithms.
   • Example: $\log_{10}\left(\frac{1000}{10}\right) = \log_{10} 1000 - \log_{10} 10 = 3 - 1 = 2$
   • Verification: $\frac{1000}{10} = 100 = 10^2$, so $\log_{10} 100 = 2$ ✓

3. Power Rule: $\log_b(M^p) = p \log_b M$

   • The logarithm of a power equals the power times the logarithm of the base.
   • Example: $\log_3(9^2) = 2 \log_3 9 = 2 \times 2 = 4$
   • Verification: $9^2 = 81 = 3^4$, so $\log_3 81 = 4$ ✓

4. Change of Base Formula: $\log_b x = \frac{\log_c x}{\log_c b} = \frac{\ln x}{\ln b}$

   • Allows conversion between different bases using natural log or common log.
   • Example: $\log_2 16 = \frac{\ln 16}{\ln 2} = \frac{2.773}{0.693} \approx 4$
   • Verification: $2^4 = 16$ ✓

5. Special Values:

   • $\log_b 1 = 0$ (because $b^0 = 1$)
   • $\log_b b = 1$ (because $b^1 = b$)
   • $\log_b b^x = x$ (because $b^x = b^x$)
   • $b^{\log_b x} = x$ (inverse relationship)

Exponential Functions:

Exponential growth and decay follow the form $A(t) = A_0 e^{kt}$, where:

• $A_0$ is the initial amount
• $k$ is the growth rate (positive for growth, negative for decay)
• $t$ is time
• $e \approx 2.71828$ is Euler's number

Alternatively: $A(t) = A_0 b^t$ where $b = 1 + r$ ($r$ is the rate of change).

Worked Example 3: Expand $\log_3\left(\frac{27x^2}{y}\right)$ using logarithmic laws

Step 1: Apply the quotient rule to separate numerator and denominator:

Step 2: Apply the product rule to the first term:

Step 3: Evaluate $\log_3 27$ and apply the power rule:

• $27 = 3^3$, so $\log_3 27 = 3$
• $\log_3(x^2) = 2\log_3 x$

Step 4: Combine:

Worked Example 4: Solve $2^{x+1} = 8$

Step 1: Express both sides as powers of the same base:

• $2^{x+1} = 8$
• $8 = 2^3$
• Therefore: $2^{x+1} = 2^3$

Step 2: Equate exponents (since the bases are equal):

Step 3: Solve for $x$:

Verification: $2^{2+1} = 2^3 = 8$ ✓

Alternative Method Using Logarithms:

Step 1: Take the natural log of both sides:

Step 2: Apply the power rule:

Step 3: Solve for $x$:

Worked Example 5: A concrete sample undergoes exponential decay of compressive strength due to sulfate attack. If initial strength is 40 MPa and strength decreases by 5% annually, find the strength after 4 years.

Step 1: Identify the exponential decay model:

• Initial strength: $A_0 = 40$ MPa
• Annual decay rate: 5% means $r = -0.05$
• Strength at time $t$ (years): $A(t) = A_0(1 + r)^t = 40(1 - 0.05)^t = 40(0.95)^t$

Step 2: Calculate strength after 4 years:

Step 3: Compute $(0.95)^4$:

Step 4: Final answer:

Verification using logarithms: We could also express this as $A(t) = 40 e^{-0.0513t}$ where $\ln(0.95) \approx -0.0513$, giving $A(4) \approx 40 e^{-0.2052} \approx 32.6$ MPa.

Quadratic equations appear throughout engineering practice—from parabolic arch designs in NSCP 2015 to deflection curves and optimization problems. A quadratic equation has the standard form:

where $a$, $b$, and $c$ are constants and $x$ is the variable.

The Quadratic Formula:

The roots (solutions) of any quadratic equation are given by:

This formula gives up to two solutions corresponding to the $\pm$ sign.

The Discriminant:

The discriminant is $\Delta = b^2 - 4ac$. It determines the nature and number of roots:

• $\Delta > 0$ (positive): Two distinct real roots

  • $x_1 = \frac{-b + \sqrt{\Delta}}{2a}$ and $x_2 = \frac{-b - \sqrt{\Delta}}{2a}$
  • The parabola crosses the $x$-axis at two points.
  • Example: $x^2 - 5x + 6 = 0$ → $\Delta = 25 - 24 = 1 > 0$ → two distinct roots

• $\Delta = 0$ (zero): One repeated real root (or a double root)

  • $x = \frac{-b}{2a}$ (a single value)
  • The parabola touches the $x$-axis at exactly one point (vertex).
  • Example: $x^2 - 4x + 4 = 0$ → $\Delta = 16 - 16 = 0$ → $x = 2$ (double root)

• $\Delta < 0$ (negative): Two complex conjugate roots

  • $x = \frac{-b \pm i\sqrt{|\Delta|}}{2a}$ (involving the imaginary unit $i$)
  • The parabola does not cross the $x$-axis (remains entirely above or below it).
  • These roots are not used in most engineering contexts requiring real physical solutions.
  • Example: $x^2 + x + 1 = 0$ → $\Delta = 1 - 4 = -3 < 0$ → complex roots

Vieta's Formulas (Relationships Between Roots and Coefficients):

For a quadratic $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2$:

• Sum of roots: $x_1 + x_2 = -\frac{b}{a}$
• Product of roots: $x_1 \cdot x_2 = \frac{c}{a}$

These relationships are useful for constructing equations or verifying solutions.

Factoring Method:

If the quadratic can be factored, it may be solved more quickly:

This gives roots $x_1$ and $x_2$ immediately.

Worked Example 6: Solve $x^2 - 5x + 6 = 0$ using both factoring and the quadratic formula

Method 1: Factoring

Step 1: Look for two numbers that multiply to $c = 6$ and add to $b = -5$.

• Candidates: $(−1, −6)$, $(−2, −3)$, $(−3, −2)$, $(−6, −1)$
• Check $(−2, −3)$: $(-2) + (-3) = -5$ ✓ and $(-2) \times (-3) = 6$ ✓

Step 2: Factor:

Step 3: Solve:

Method 2: Quadratic Formula

Step 1: Identify coefficients: $a = 1$, $b = -5$, $c = 6$

Step 2: Calculate the discriminant:

Step 3: Apply the quadratic formula:

Step 4: Compute the two roots:

Verification using Vieta's formulas:

• Sum: $2 + 3 = 5 = -(-5)/1$ ✓
• Product: $2 \times 3 = 6 = 6/1$ ✓

Worked Example 7: Solve $2x^2 + 3x - 5 = 0$

Step 1: Identify coefficients: $a = 2$, $b = 3$, $c = -5$

Step 2: Calculate the discriminant:

Step 3: Since $\Delta = 49 > 0$, there are two distinct real roots. Apply the quadratic formula:

Step 4: Compute the roots:

Verification:

• For $x = 1$: $2(1)^2 + 3(1) - 5 = 2 + 3 - 5 = 0$ ✓
• For $x = -5/2$: $2(-5/2)^2 + 3(-5/2) - 5 = 2(25/4) - 15/2 - 5 = 25/2 - 15/2 - 10/2 = 0$ ✓

Worked Example 8: Find the condition on $m$ such that $x^2 - (m+1)x + m = 0$ has real roots

Step 1: For real roots, the discriminant must be non-negative: $\Delta \geq 0$

Step 2: Set up the discriminant inequality:

Step 3: Expand and simplify:

Step 4: Interpret the result:
Since $(m - 1)^2$ is always non-negative (a perfect square), the inequality holds for all real values of $m$. Therefore, the equation always has real roots.

Special case: When $m = 1$, $(m-1)^2 = 0$, giving a repeated root.

Polynomial Equations (Higher Degree):

For polynomials of degree $n$:

Solving cubic and quartic equations can be complex, but several strategies help:

1. Rational Root Theorem: If $p/q$ is a rational root (in lowest terms), then $p$ divides $a_0$ and $q$ divides $a_n$.
2. Factor by grouping: Rearrange and group terms to factor out common expressions.
3. Synthetic division or polynomial long division: Once one root is found, divide to reduce the degree.
4. Special forms: Recognize patterns like difference of squares, sum/difference of cubes, etc.

Worked Example 9: Solve $x^3 - 6x^2 + 11x - 6 = 0$

Step 1: Apply the Rational Root Theorem. Possible rational roots are divisors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$

Step 2: Test $x = 1$:
$$1 - 6 + 11 - 6 = 0$$ ✓
So $x = 1$ is a root, and $(x - 1)$ is a factor.

Step 3: Perform synthetic division or polynomial long division to find the quotient when dividing by $(x - 1)$:

Step 4: Factor the resulting quadratic:

Step 5: Complete factorization:

Step 6: Solutions:

Progressions (sequences with specific patterns) are common in engineering analysis. Arithmetic progressions model linear change (constant difference), while geometric progressions model exponential growth or decay (constant ratio). Both appear in loan calculations, stress distributions, and sampling designs.

Arithmetic Progression (AP):

An AP is a sequence where the difference between consecutive terms is constant. This constant is called the common difference, denoted $d$.

General sequence: $a_1, a_2, a_3, \ldots, a_n$ where $a_n - a_{n-1} = d$ for all $n \geq 2$.

Formula for the $n$-th term:

where:

• $a_n$ is the $n$-th term
• $a_1$ is the first term
• $d$ is the common difference
• $n$ is the term number

Sum of the first $n$ terms of an AP:

Both forms are useful: use the first when you know $a_1$ and $d$; use the second when you know both endpoints.

Worked Example 10: AP with known first term and common difference

Find the sum of the first 15 terms of the AP: $3, 7, 11, 15, \ldots$

Step 1: Identify the parameters:

• First term: $a_1 = 3$
• Common difference: $d = 7 - 3 = 4$
• Number of terms: $n = 15$

Step 2: Find the 15th term using $a_n = a_1 + (n-1)d$:

Step 3: Calculate the sum:

Verification using the alternative formula:
$$S_{15} = \frac{15}{2}[2(3) + (15-1)(4)] = \frac{15}{2}[6 + 56] = \frac{15}{2}(62) = 465$$ ✓

Worked Example 11: Finding a specific term in an AP

In an AP, the 5th term is 13 and the 12th term is 34. Find the first term and the common difference.

Step 1: Set up equations using $a_n = a_1 + (n-1)d$:

• For $n = 5$: $a_5 = a_1 + 4d = 13$ ... (1)
• For $n = 12$: $a_{12} = a_1 + 11d = 34$ ... (2)

Step 2: Subtract equation (1) from equation (2):

Step 3: Substitute $d = 3$ into equation (1):

Verification:

• $a_5 = 1 + 4(3) = 13$ ✓
• $a_{12} = 1 + 11(3) = 34$ ✓
• The AP is: $1, 4, 7, 10, 13, \ldots$

Geometric Progression (GP):

A GP is a sequence where the ratio between consecutive terms is constant. This constant is called the common ratio, denoted $r$.

General sequence: $a_1, a_2, a_3, \ldots, a_n$ where $\frac{a_n}{a_{n-1}} = r$ for all $n \geq 2$.

Formula for the $n$-th term:

where:

• $a_n$ is the $n$-th term
• $a_1$ is the first term
• $r$ is the common ratio
• $n$ is the term number

Sum of the first $n$ terms of a GP:

Alternatively (for $r \neq 1$):

Both forms are equivalent; choose based on convenience.

Special case: $r = 1$
When $r = 1$, all terms are equal to $a_1$, so $S_n = n \cdot a_1$.

Sum of an Infinite GP (when $|r| < 1$):

If $|r| < 1$ (the common ratio is strictly between $-1$ and $1$), the infinite series converges:

This formula is critical in engineering for calculating present values, damping ratios, and convergent stresses.

If $|r| \geq 1$, the infinite series diverges (does not converge to a finite sum).

Worked Example 12: Find the sum of the first 8 terms of the GP: $2, 6, 18, 54, \ldots$

Step 1: Identify parameters:

• First term: $a_1 = 2$
• Common ratio: $r = 6/2 = 3$
• Number of terms: $n = 8$

Step 2: Since $r \neq 1$, use the GP sum formula:

Verification (compute individual terms):

• $a_1 = 2$
• $a_2 = 2 \times 3 = 6$
• $a_3 = 2 \times 3^2 = 18$
• $a_4 = 2 \times 3^3 = 54$
• $a_5 = 2 \times 3^4 = 162$
• $a_6 = 2 \times 3^5 = 486$
• $a_7 = 2 \times 3^6 = 1458$
• $a_8 = 2 \times 3^7 = 4374$
• Sum: $2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 = 6560$ ✓

Worked Example 13: Find the sum of an infinite GP

Find the sum of the infinite series: $8 + 4 + 2 + 1 + 0.5 + \ldots$

Step 1: Identify parameters:

• First term: $a_1 = 8$
• Common ratio: $r = 4/8 = 1/2 = 0.5$
• Check: $|r| = 0.5 < 1$ ✓ (series converges)

Step 2: Apply the infinite sum formula:

Interpretation: As we add more and more terms, their sum approaches (but never exceeds) 16.

Worked Example 14: Engineering application—Damping in structures

A vibrating beam experiences damping such that each oscillation amplitude is 80% of the previous amplitude. If the initial amplitude is 10 mm, what is the total displacement (sum of all oscillations) theoretically?

Step 1: Identify the GP:

• First amplitude: $a_1 = 10$ mm
• Common ratio: $r = 0.8$ (each is 80% of previous)
• Check: $|r| = 0.8 < 1$ ✓ (infinite sum converges)

Step 2: Calculate the theoretical total displacement:

Interpretation: Although the beam oscillates infinitely, the total displacement from all oscillations converges to 50 mm. This is crucial in predicting fatigue damage and stress concentration factors.

The binomial theorem provides a formula for expanding expressions of the form $(a + b)^n$ without multiplying them out term by term. This is invaluable for approximations in engineering (e.g., small deformation theory) and combinatorial counting.

The Binomial Theorem (General Form):

where $\binom{n}{k}$ (read "$n$ choose $k$") is the binomial coefficient:

Here, $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$ is the factorial.

Expanded form:

Key Properties of Binomial Coefficients:

1. $\binom{n}{0} = 1$ and $\binom{n}{n} = 1$
2. $\binom{n}{k} = \binom{n}{n-k}$ (symmetry)
3. $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ (Pascal's identity)

Pascal's Triangle:

Binomial coefficients can be arranged in Pascal's triangle:

n=0:                    1
n=1:                  1   1
n=2:                1   2   1
n=3:              1   3   3   1
n=4:            1   4   6   4   1
n=5:          1   5  10  10   5   1
n=6:        1   6  15  20  15   6   1

Each entry is the sum of the two entries above it. This allows rapid lookup of binomial coefficients without calculating factorials.

General Term ($(r+1)$-th term):

The $(r+1)$-th term in the expansion of $(a+b)^n$ is:

where $r = 0, 1, 2, \ldots, n$. (Note: we count from $r=0$, so the "first" term has $r=0$.)

Worked Example 15: Expand $(2x + 3)^4$

Step 1: Identify $a = 2x$, $b = 3$, $n = 4$

Step 2: Write the general expansion:

Step 3: Calculate binomial coefficients using Pascal's triangle (or factorials):

• $\binom{4}{0} = 1$
• $\binom{4}{1} = 4$
• $\binom{4}{2} = 6$
• $\binom{4}{3} = 4$
• $\binom{4}{4} = 1$

Step 4: Compute each term:

• $k=0$: $\binom{4}{0}(2x)^4(3)^0 = 1 \times 16x^4 \times 1 = 16x^4$
• $k=1$: $\binom{4}{1}(2x)^3(3)^1 = 4 \times 8x^3 \times 3 = 96x^3$
• $k=2$: $\binom{4}{2}(2x)^2(3)^2 = 6 \times 4x^2 \times 9 = 216x^2$
• $k=3$: $\binom{4}{3}(2x)^1(3)^3 = 4 \times 2x \times 27 = 216x$
• $k=4$: $\binom{4}{4}(2x)^0(3)^4 = 1 \times 1 \times 81 = 81$