7.3 Logarithmic Properties and Solving Exponential/Logarithmic Equations

7.3 Logarithmic Properties and Solving Exponential/Logarithmic Equations

Quick Answer: A logarithm is the inverse operation of exponentiation: $\log_b(y) = x \iff b^x = y$, where base $b > 0, b \neq 1$, and argument $y > 0$. Common logarithms have base 10 ($\log x$), while natural logarithms have base $e$ ($\ln x$). Logarithms obey three operational rules: Product Rule $\log_b(xy) = \log_b x + \log_b y$, Quotient Rule $\log_b(x/y) = \log_b x - \log_b y$, and Power Rule $\log_b(x^k) = k \log_b x$. The Change of Base formula is $\log_b a = \frac{\ln a}{\ln b}$. Exponential equations are solved by isolating the power and taking the natural/common log of both sides. Logarithmic equations are solved by condensing terms and exponentiating, always testing for extraneous solutions that make any argument non-positive ($u \le 0$).

---

Definition of the Logarithmic Function (AII-F.BF.5a, AII-F.LE.4)

The fundamental axiom to memorize in Algebra II is: a logarithm is an exponent. The logarithmic equation $\log_b(y) = x$ answers the question: "To what exponent $x$ must the base $b$ be raised to yield the argument $y$?"

$$\log_b(y) = x \iff b^x = y$$

• Base Restrictions: The base $b$ must be strictly positive and cannot equal one ($b > 0, b \neq 1$).
• Argument Domain: Because positive bases raised to any real power always produce positive outputs ($b^x > 0$), the argument $y$ must be strictly positive ($y > 0$). You cannot take the logarithm of zero or a negative number in the real number system.

Graphing the Logarithmic Function

As the inverse of $f(x) = b^x$, the logarithmic curve $g(x) = \log_b(x)$ reflects $b^x$ across the line $y = x$:

• $x$-Intercept: $(1, 0)$, because $b^0 = 1 \iff \log_b(1) = 0$.
• Vertical Asymptote: The line $x = 0$ (the $y$-axis). As $x \to 0^+$, $y \to -\infty$ (for $b > 1$).
• Domain & Range: Domain is $(0, \infty)$; Range is $(-\infty, \infty)$.

Special Logarithmic Bases

1. Common Logarithm (Base 10): When written without a subscript, $\log(x)$ implies base 10: $\log(x) = \log_{10}(x)$.
2. Natural Logarithm (Base $e$): Logarithms to Euler's base $e$ are denoted as $\ln(x)$: $\ln(x) = \log_e(x)$. In particular, $\ln(e) = 1$ and $\ln(1) = 0$.

---

Master Properties of Logarithms

Because logarithms are exponents, every logarithmic law corresponds directly to an exponential law:

1. The Product Rule

The logarithm of a product is the sum of the individual logarithms:

$$\log_b(M \cdot N) = \log_b(M) + \log_b(N)$$

Proof connection: When multiplying like bases, add exponents: $b^u \cdot b^v = b^{u+v}$.

2. The Quotient Rule

The logarithm of a quotient is the difference of the logarithms:

$$\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$$

Proof connection: When dividing like bases, subtract exponents: $\frac{b^u}{b^v} = b^{u-v}$.

3. The Power Rule

The logarithm of a power equals the exponent multiplied by the logarithm of the base:

$$\log_b(M^k) = k \cdot \log_b(M)$$

Proof connection: When raising a power to a power, multiply exponents: $(b^u)^k = b^{uk}$.

4. Inverse and Identity Properties

• $\log_b(b^x) = x$ and $b^{\log_b(x)} = x$ (for $x > 0$). In particular, $\ln(e^x) = x$ and $e^{\ln(x)} = x$.
• $\log_b(b) = 1$, since $b^1 = b$.
• $\log_b(1) = 0$, since $b^0 = 1$.

5. Change of Base Formula

To compute logarithms with arbitrary bases on standard calculators, rewrite the logarithm as a ratio of natural or common logs:

$$\log_b(a) = \frac{\ln(a)}{\ln(b)} = \frac{\log(a)}{\log(b)}$$

---

Expanding and Condensing Logarithmic Expressions

• Expanding: Breaks a single complex logarithmic argument into a linear combination of simpler logs. Apply the quotient rule first, followed by the product rule, and finally the power rule to bring exponents to the front as coefficients.
• Condensing: Combines multiple logarithmic terms with identical bases into a single compact logarithm. Always apply the power rule in reverse first (moving coefficients into exponents), then apply the product and quotient rules from left to right.

---

Solving Exponential Equations Algebraically (AII-F.LE.4)

To solve an exponential equation where the unknown variable resides in the exponent, follow this four-step algebraic method:

1. Isolate the Power: Move all constants and leading coefficients away from the base expression $b^{g(x)}$.
2. Take the Logarithm of Both Sides: Apply the natural logarithm ($\ln$) or common logarithm ($\log$) to both sides.
3. Apply the Power Rule: Pull the exponent variable down out of the power: $\ln(b^{g(x)}) = g(x) \cdot \ln(b)$. If the base is $e$, simplify directly using $\ln(e^u) = u$.
4. Solve for the Variable: Divide by $\ln(b)$ and perform standard linear algebra to isolate $x$.

Example: Solving Base $e$ Exponential Models

Solve $500e^{0.04t} = 900$ algebraically:

1. Isolate the exponential: $e^{0.04t} = \frac{900}{500} = 1.8$.
2. Take the natural logarithm: $\ln(e^{0.04t}) = \ln(1.8)$.
3. Simplify the left side: $0.04t = \ln(1.8)$.
4. Divide by 0.04: $t = \frac{\ln(1.8)}{0.04} \approx \frac{0.587787}{0.04} \approx 14.695$.

---

Solving Logarithmic Equations & The Extraneous Root Mandate (AII-A.REI.2)

Logarithmic equations fall into two primary structural forms:

Form 1: Equating Arguments (One-to-One Property)

$$\log_b(f(x)) = \log_b(g(x)) \implies f(x) = g(x)$$

Condense each side into a single logarithm first. Once each side consists of a single logarithm with identical base, set their arguments equal.

Form 2: Converting to Exponential Form

$$\log_b(f(x)) = c \implies f(x) = b^c$$

Condense all logarithmic terms on one side into a single logarithm, then rewrite the equation in exponential form.

The Mandatory Extraneous Solution Check

Because the real-number domain of $\log_b(u)$ requires $u > 0$, algebraic procedures like exponentiation can introduce candidate solutions that fail in the original equation.

---

Summary of Logarithmic Rules and Operations

---

Worked Examples

Worked Problem 1: Solving an Exponential Equation Algebraically

Problem: Algebraically solve the equation $250(1.05)^t = 800$ for $t$. Express your answer as an exact value using natural logarithms, and round to the nearest hundredth.

• Step 1: Isolate the exponential base. Divide both sides by $250$:

  $$(1.05)^t = \frac{800}{250} = 3.2$$

• Step 2: Apply the natural logarithm to both sides.

  $$\ln\left((1.05)^t\right) = \ln(3.2)$$

• Step 3: Apply the power rule.

  $$t \cdot \ln(1.05) = \ln(3.2)$$

• Step 4: Solve for $t$ exactly.

  $$t = \frac{\ln(3.2)}{\ln(1.05)}$$

• Step 5: Compute the decimal approximation.

  $$t \approx \frac{1.163151}{0.048790} \approx 23.8399 \implies t \approx 23.84$$

Worked Problem 2: Condensing a Multi-Term Logarithmic Expression

Problem: Express $2\log_3(x) + \frac{1}{2}\log_3(x + 4) - \log_3(x^2 - 1)$ as a single logarithm with a coefficient of 1.

• Step 1: Apply the power rule to move all coefficients to exponents.

  $$\log_3(x^2) + \log_3\left((x + 4)^{\frac{1}{2}}\right) - \log_3(x^2 - 1)$$

  Rewrite fractional exponent as a square root:

  $$\log_3(x^2) + \log_3(\sqrt{x + 4}) - \log_3(x^2 - 1)$$

• Step 2: Apply the product rule to positive terms.

  $$\log_3\left(x^2 \sqrt{x + 4}\right) - \log_3(x^2 - 1)$$

• Step 3: Apply the quotient rule for the subtracted term.

  $$\log_3\left(\frac{x^2 \sqrt{x + 4}}{x^2 - 1}\right)$$

Worked Problem 3: Solving a Logarithmic Equation with Extraneous Root Elimination

Problem: Algebraically solve $\log_2(x + 3) + \log_2(x - 1) = 5$ for $x$.

• Step 1: Condense the left side using the product rule.

  $$\log_2[(x + 3)(x - 1)] = 5$$

• Step 2: Convert to equivalent exponential form.

  $$(x + 3)(x - 1) = 2^5 = 32$$

• Step 3: Expand and set the quadratic equal to zero.

  $$x^2 + 2x - 3 = 32 \implies x^2 + 2x - 35 = 0$$

• Step 4: Factor the quadratic.

  $$(x + 7)(x - 5) = 0 \implies x = -7 \quad \text{or} \quad x = 5$$

• Step 5: Check candidate roots in original arguments.

  • Test $x = -7$: $$\log_2(-7 + 3) = \log_2(-4) \quad \text{and} \quad \log_2(-7 - 1) = \log_2(-8)$$ Because arguments cannot be negative, $x = -7$ is extraneous and must be rejected.
  • Test $x = 5$: $$\log_2(5 + 3) + \log_2(5 - 1) = \log_2(8) + \log_2(4) = 3 + 2 = 5$$ This is true and valid.

• Step 6: State the final solution.

  $$\{5\}$$

---

Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Distributing logarithms across addition. Students often write $\log(a + b) = \log a + \log b$. This is completely false. Logarithms turn products into sums ($\log(ab) = \log a + \log b$), but $\log(a + b)$ cannot be expanded algebraically.
• Pitfall 2: Confusing $\frac{\log a}{\log b}$ with $\log\left(\frac{a}{b}\right)$. The change of base formula is the quotient of two logarithms: $\frac{\log a}{\log b} = \log_b a$. The quotient rule is the logarithm of a single fraction: $\log\left(\frac{a}{b}\right) = \log a - \log b$.
• Pitfall 3: Failing to show the extraneous root rejection. On constructed-response items, you must explicitly show or state why an extraneous solution is rejected. Simply omitting the negative root without notation may result in a 1-credit loss. Write: "$x = -7$ is rejected because the argument of a logarithm must be positive."