3.3 Exponential Functions, Properties & Equations

3.3 Exponential Functions, Properties & Equations

Exponential functions model dynamic phenomena in economics and finance, including compound interest, capital accumulation, population dynamics, and continuous depreciation. On the Bocconi Admission Test, mastery of exponent laws, algebraic substitutions, and base-dependent inequality rules is essential for completing numerical questions efficiently without a calculator.

---

1. Definition and Foundational Properties of $f(x) = a^x$

An exponential function is a function of the form:

$$f(x) = a^x$$

where the base $a$ is a fixed positive constant different from 1: $a \in (0, 1) \cup (1, \infty)$.

Why Are Bases Restricted to $a > 0$ and $a \neq 1$?

• If $a < 0$: Powers with rational exponents would yield non-real numbers (e.g., $(-4)^{1/2} = \sqrt{-4} = 2i \notin \mathbb{R}$). An unbroken continuous curve in $\mathbb{R}$ cannot be constructed with a negative base.
• If $a = 0$: The expression $0^x = 0$ for all $x > 0$, but is undefined for $x \le 0$ ($0^0$ is indeterminate, $0^{-1} = 1/0$ is undefined).
• If $a = 1$: The expression $1^x = 1$ collapses to the constant horizontal line $y = 1$, losing all exponential characteristics.

Analytical Characteristics of $y = a^x$

1. Domain: $\text{Dom}(f) = \mathbb{R} = (-\infty, \infty)$.
2. Range: $\text{Ran}(f) = (0, \infty)$. The function is strictly positive: $a^x > 0$ for all $x \in \mathbb{R}$. It never touches the $x$-axis.
3. Universal Intercept: Because $a^0 = 1$ for all valid bases, every exponential curve passes through the $y$-intercept $(0, 1)$.
4. Horizontal Asymptote: The line $y = 0$ (the $x$-axis) is a horizontal asymptote for all exponential functions.
5. Bijectivity: Exponential functions $f: \mathbb{R} \to (0, \infty)$ are strictly monotonic and therefore bijective, ensuring they possess two-sided inverses (logarithms).

---

2. Monotonicity Regimes: Growth vs. Decay

The qualitative behavior of $y = a^x$ is determined entirely by the magnitude of its base $a$:

Regime 1: $a > 1$ (Exponential Growth)

• The function is strictly increasing across $\mathbb{R}$: $$x_1 < x_2 \iff a^{x_1} < a^{x_2}$$
• Asymptotic behavior: $$\lim_{x \to \infty} a^x = \infty \qquad \text{and} \qquad \lim_{x \to -\infty} a^x = 0^+ \quad (\text{horizontal asymptote as } x \to -\infty)$$
• Relative steepness: For bases $b > a > 1$, $b^x > a^x$ when $x > 0$, whereas $b^x < a^x$ when $x < 0$.

Regime 2: $0 < a < 1$ (Exponential Decay)

• The function is strictly decreasing across $\mathbb{R}$: $$x_1 < x_2 \iff a^{x_1} > a^{x_2}$$
• Decay relationship: Setting $b = \frac{1}{a} > 1$, we observe: $$a^x = \left(\frac{1}{b}\right)^x = b^{-x}$$ Thus, the graph of an exponential decay function is the exact geometric reflection of the growth function $y = b^x$ across the $y$-axis ($x \mapsto -x$).
• Asymptotic behavior: $$\lim_{x \to \infty} a^x = 0^+ \quad (\text{horizontal asymptote as } x \to \infty) \qquad \text{and} \qquad \lim_{x \to -\infty} a^x = \infty$$

---

3. The Natural Exponential $e^x$ & Continuous Growth

In economic modeling, interest compounded continuously and exponential processes are expressed using the natural base $e$ (Euler's number):

$$\ne = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.718281828...$$

Compound Interest Transition

When a principal $P$ earns an annual nominal interest rate $r$ compounded $n$ times per year for $t$ years, the accumulated balance is:

$$A(t) = P \left(1 + \frac{r}{n}\right)^{nt}$$

Taking the limit as compounding becomes continuous ($n \to \infty$):

$$\lim_{n \to \infty} P \left(1 + \frac{r}{n}\right)^{nt} = P \cdot \left[\lim_{n/r \to \infty} \left(1 + \frac{1}{n/r}\right)^{n/r}\right]^{rt} = P e^{rt}$$

---

4. Master Table of Operational Exponent Laws

To simplify expressions rapidly during the exam, memorize and deploy these algebraic rules:

Critical Bocconi Trap: Exponents do NOT distribute across addition: $(a + b)^x \neq a^x + b^x$. Furthermore, powers of powers are not associative: $a^{(b^c)} \neq (a^b)^c$. For example, $2^{(3^2)} = 2^9 = 512$, whereas $(2^3)^2 = 2^6 = 64$.

---

5. Solving Exponential Equations

Exponential equations on the Bocconi test fall into two primary structures:

Method 1: Equating Bases

When both sides of an equation can be rewritten as powers of a single common base $a$:

$$a^{f(x)} = a^{g(x)} \iff f(x) = g(x)$$

Worked Example: Solve for $x \in \mathbb{R}$:

$$8^{x^2 - 1} = 16^{x + 1}$$

Express both bases as powers of 2 ($8 = 2^3$, $16 = 2^4$):

$$(2^3)^{x^2 - 1} = (2^4)^{x + 1} \implies 2^{3(x^2 - 1)} = 2^{4(x + 1)}$$

Equating exponents:

$$3x^2 - 3 = 4x + 4 \implies 3x^2 - 4x - 7 = 0$$

Factoring the quadratic:

$$(3x - 7)(x + 1) = 0 \implies x = \frac{7}{3} \quad \text{or} \quad x = -1$$

Method 2: Quadratic Forms via Substitution ($u = a^x$)

Equations containing linear combinations of $a^{2x} = (a^x)^2$ and $a^x$ can be transformed into polynomials by substituting $u = a^x$. Crucial constraint: Because $a^x > 0$ for all real $x$, candidates must enforce:

$$\nu > 0$$

Any candidate root where $u \le 0$ must be immediately rejected as extraneous.

Worked Example: Solve for $x \in \mathbb{R}$:

$$4^{x} - 5 \cdot 2^{x+1} + 16 = 0$$

Deconstruct the exponential terms into powers of $2^x$:

• $4^x = (2^2)^x = (2^x)^2$
• $2^{x+1} = 2^x \cdot 2^1 = 2 \cdot 2^x$, so $5 \cdot 2^{x+1} = 10 \cdot 2^x$

The equation becomes:

$$(2^x)^2 - 10(2^x) + 16 = 0$$

Substitute $u = 2^x > 0$:

$$\nu^2 - 10u + 16 = 0 \implies (u - 2)(u - 8) = 0 \implies u = 2 \quad \text{or} \quad u = 8$$

Back-substitute to find $x$:

1. $2^x = 2 = 2^1 \implies x = 1$
2. $2^x = 8 = 2^3 \implies x = 3$

Both values are valid real solutions: $x \in \{1, 3\}$.

---

6. Solving Exponential Inequalities: The Direction Reversal Rule

When solving inequalities of the form $a^{f(x)} < a^{g(x)}$, the direction of the inequality symbol depends entirely on whether the base is greater than 1 or between 0 and 1:

$$\begin{cases} f(x) < g(x) & \text{if } a > 1 \quad (\text{preserves direction}) \\ f(x) > g(x) & \text{if } 0 < a < 1 \quad (\text{reverses direction}) \end{cases}$$

Worked Example: Fractional Base Inequality

Solve for $x \in \mathbb{R}$:

$$\left(\frac{1}{2}\right)^{x^2 - 3} \le \left(\frac{1}{4}\right)^{x}$$

Step 1: Unify the Base. Rewrite $1/4 = (1/2)^2$:

$$\left(\frac{1}{2}\right)^{x^2 - 3} \le \left(\left(\frac{1}{2}\right)^2\right)^x \implies \left(\frac{1}{2}\right)^{x^2 - 3} \le \left(\frac{1}{2}\right)^{2x}$$

Step 2: Apply Monotonicity. Because the base $a = 1/2$ satisfies $0 < a < 1$, the function is strictly decreasing. Comparing exponents requires reversing the inequality sign:

$$x^2 - 3 \ge 2x \iff x^2 - 2x - 3 \ge 0$$

Step 3: Solve the Quadratic Inequality.

$$(x - 3)(x + 1) \ge 0$$

The roots are $x = -1$ and $x = 3$. The quadratic is positive outside the roots:

$$x \in (-\infty, -1] \cup [3, \infty)$$

Alternative Verification: Converting to base 2 yields $2^{-(x^2 - 3)} \le 2^{-2x} \implies 3 - x^2 \le -2x \implies x^2 - 2x - 3 \ge 0$, confirming the identical solution set.

---

7. Common Bocconi Traps & Exam Strategies

1. Retaining Negative Substitution Roots: In $a^{2x} + b a^x + c = 0$, if solving for $u = a^x$ yields $u_1 = 4$ and $u_2 = -2$, students who write $2^x = -2 \implies x = \text{undefined}$ sometimes forget to discard $u_2$, or mistakenly conclude there are no solutions at all. Always discard only the non-positive $u$-values.
2. Forgetting to Reverse Signs: Failing to invert the inequality when the base is fractional ($0 < a < 1$, such as $0.5, 1/3, 0.2$) is the single most common student error in Bocconi exponential questions.
3. Base Dependent Expressions: Equations like $x^{x^2 - 1} = x^3$ have variable bases and require checking each degenerate base separately — $x = 1$, $x = -1$ and $x = 0$ — before equating exponents for the general case $x > 0$, $x \neq 1$.