3.6 Exponential & Logarithmic Equations

3.6 Exponential & Logarithmic Equations

Exponential and logarithmic functions are inverse mathematical operations. Solving equations involving these functions requires applying logarithm laws, exponent rules, base matching, and domain restriction screening. These concepts represent high-yield content on the CLEP College Algebra exam.

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1. Solving Exponential Equations

An exponential equation features variables in exponent positions ($b^{g(x)} = c$).

Strategy A: Equating Bases (One-to-One Property)

For base $b > 0$ and $b \neq 1$:

$b^u = b^v \iff u = v$

If both sides can be expressed using a common base $b$, equate exponents to solve.

Worked Example 1: Equating Bases

Solve $8^{x - 1} = 16^{2x + 3}$.

1. Express $8$ and $16$ as powers of $2$ ($8 = 2^3$ and $16 = 2^4$): $(2^3)^{x - 1} = (2^4)^{2x + 3} \implies 2^{3x - 3} = 2^{8x + 12}$
2. Equate exponents: $3x - 3 = 8x + 12 \implies -5x = 15 \implies x = -3$ Solution: $x = -3$.

Strategy B: Taking Natural Logarithms of Both Sides

When bases cannot be matched, take the natural logarithm ($\ln$) of both sides and apply the Power Property ($\ln(a^k) = k \ln a$).

Worked Example 2: Taking Natural Logs

Solve $3^{2x - 1} = 7^{x + 2}$.

1. Take $\ln$ of both sides: $\ln(3^{2x - 1}) = \ln(7^{x + 2}) \implies (2x - 1)\ln 3 = (x + 2)\ln 7$
2. Distribute and isolate $x$: $2x \ln 3 - \ln 3 = x \ln 7 + 2 \ln 7$ $2x \ln 3 - x \ln 7 = \ln 3 + 2 \ln 7$ $x(2 \ln 3 - \ln 7) = \ln 3 + 2 \ln 7 \implies x = \frac{\ln 3 + 2 \ln 7}{2 \ln 3 - \ln 7}$

Strategy C: Quadratic-in-Form Exponential Equations

Equations such as $e^{2x} - 5e^x + 6 = 0$ can be solved by substituting $u = e^x$, yielding $u^2 - 5u + 6 = 0 \implies (u - 2)(u - 3) = 0$. Thus $e^x = 2 \implies x = \ln 2$, and $e^x = 3 \implies x = \ln 3$.

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2. Solving Logarithmic Equations & Domain Restrictions

A logarithmic equation contains variable logarithmic arguments.

Fundamental Logarithmic Laws

• Inverse Definition: $\log_b(u) = c \iff u = b^c$
• Product Rule: $\log_b(uv) = \log_b(u) + \log_b(v)$
• Quotient Rule: $\log_b\left(\frac{u}{v}\right) = \log_b(u) - \log_b(v)$
• Power Rule: $\log_b(u^k) = k \log_b(u)$
• Change of Base: $\log_b(u) = \frac{\ln u}{\ln b}$

Crucial Rule: Domain Screening ($u > 0$)

CRITICAL RULE: Logarithm functions $y = \log_b(u)$ are defined ONLY for strictly positive arguments ($u > 0$). You MUST substitute candidate solutions into the original log expressions. Candidates causing non-positive arguments are extraneous solutions and must be eliminated.

Worked Example 3: Logarithmic Equation with Extraneous Root

Solve $\log_3(x + 2) + \log_3(x - 4) = 3$.

1. Condense using Product Rule: $\log_3[(x + 2)(x - 4)] = 3$
2. Convert to exponential form: $(x + 2)(x - 4) = 3^3 = 27$
3. Expand and solve quadratic: $x^2 - 2x - 8 = 27 \implies x^2 - 2x - 35 = 0 \implies (x - 7)(x + 5) = 0$ Roots: $x = 7$ or $x = -5$.
4. Screen against Domain Restrictions ($u > 0$):
   • Check $x = -5$: $\log_3(-5 + 2) = \log_3(-3)$ is undefined. Extraneous.
   • Check $x = 7$: $\log_3(9) + \log_3(3) = 2 + 1 = 3$. Valid! Solution: $x = 7$.

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3. Continuous Growth, Decay & Financial Models

• Continuous Compounding Model: $A = P e^{rt}$
• Doubling Time Derivation: Set $A = 2P \implies 2P = P e^{rt} \implies 2 = e^{rt} \implies t = \frac{\ln 2}{r}$
• Radioactive Decay Model: $N(t) = N_0 e^{-kt}$, where half-life $t_{1/2} = \frac{\ln 2}{k}$.

Worked Example 4: Continuous Interest Doubling Time

An investment grows continuously at $5\%$ annual interest ($r = 0.05$). How many years will it take to double? $2 = e^{0.05 t} \implies \ln 2 = 0.05 t \implies t = \frac{\ln 2}{0.05} \approx \frac{0.693147}{0.05} \approx 13.86 \text{ years}$ Rounding to one decimal place gives $13.9$ years.

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4. Master Chapter 3 Equation Comparison Matrix