5.3 Exponential and Logarithmic Equations

5.3 Exponential and Logarithmic Equations

Solving equations involving exponential expressions and logarithms is a foundational skill evaluated on the CLEP Precalculus exam. Analytical strategies depend on the algebraic structure of the equation, ranging from equal-base matching to applying inverse logarithmic transformations and checking for extraneous roots.

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1. Solving Exponential Equations

Exponential equations fall into three primary algebraic categories:

Category A: Equations with Shared / Like Bases (One-to-One Property)

If an exponential equation can be expressed so that both sides have the same base $b > 0, b \neq 1$, apply the One-to-One Property of Exponents:

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

Worked Example 1: Solving via Shared Base Matching

Solve for $x$: $27^{2x - 1} = 9^{x + 4}$

Solution:

1. Recognize that both $27$ and $9$ are powers of base $3$: $27 = 3^3$ and $9 = 3^2$.
2. Rewrite both sides using base $3$: $(3^3)^{2x - 1} = (3^2)^{x + 4}$
3. Apply power-of-a-power rule $(b^m)^n = b^{mn}$: $3^{3(2x - 1)} = 3^{2(x + 4)} \implies 3^{6x - 3} = 3^{2x + 8}$
4. Apply the One-to-One Property ($u = v$): $6x - 3 = 2x + 8$
5. Solve the linear equation: $4x = 11 \implies x = \frac{11}{4}$

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Category B: Equations with Unlike Bases (Taking Logarithms)

When bases cannot be converted to a common integer base, take the natural logarithm ($\ln$) or common logarithm ($\log$) of both sides, then use the Power Property $\ln(b^u) = u \ln(b)$ to bring the variable down from the exponent.

Worked Example 2: Solving an Exponential Equation with Unlike Bases

Solve for $x$ exactly: $5^{2x - 3} = 3^{x + 1}$

Solution:

1. Take the natural logarithm of both sides: $\ln(5^{2x - 3}) = \ln(3^{x + 1})$
2. Apply the Power Property to bring exponents to the front: $(2x - 3)\ln(5) = (x + 1)\ln(3)$
3. Distribute $\ln(5)$ and $\ln(3)$ across the binomials: $2x\ln(5) - 3\ln(5) = x\ln(3) + \ln(3)$
4. Collect all terms containing $x$ on the left side, and constants on the right: $2x\ln(5) - x\ln(3) = 3\ln(5) + \ln(3)$
5. Factor out $x$: $x \left(2\ln(5) - \ln(3)\right) = 3\ln(5) + \ln(3)$
6. Divide to isolate $x$: $x = \frac{3\ln(5) + \ln(3)}{2\ln(5) - \ln(3)}$
7. Optionally condense using logarithmic properties: $3\ln(5) + \ln(3) = \ln(5^3 \cdot 3) = \ln(375)$ $2\ln(5) - \ln(3) = \ln\left(\frac{5^2}{3}\right) = \ln\left(\frac{25}{3}\right)$ $x = \frac{\ln(375)}{\ln(25/3)}$

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Category C: Quadratic-Form Exponential Equations

Equations of the form $a(b^x)^2 + b(b^x) + c = 0$ (or $a e^{2x} + b e^x + c = 0$) are solved by using substitution $u = b^x$, solving the quadratic equation for $u$, and then solving $b^x = u$.

Worked Example 3: Quadratic Form Exponential Equation

Solve for all real values of $x$: $e^{2x} - 5e^x + 6 = 0$

Solution:

1. Let $u = e^x$. Then $e^{2x} = (e^x)^2 = u^2$. Substitute $u$: $u^2 - 5u + 6 = 0$
2. Factor the quadratic equation: $(u - 2)(u - 3) = 0 \implies u = 2 \quad \text{or} \quad u = 3$
3. Substitute $e^x$ back in for $u$:
   • Case 1: $e^x = 2 \implies x = \ln(2)$
   • Case 2: $e^x = 3 \implies x = \ln(3)$
4. Both solutions are valid real numbers since $2 > 0$ and $3 > 0$. Solution set: ${\ln 2, , \ln 3}$.

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2. Solving Logarithmic Equations

Logarithmic equations are solved by using one of two primary strategies:

Strategy 1: Definition of Logarithm ($\log_b(u) = c \iff u = b^c$)

Isolate the logarithmic term on one side, then rewrite in exponential form.

Strategy 2: One-to-One Property ($\log_b(u) = \log_b(v) \iff u = v$)

Use logarithmic properties to condense both sides into a single logarithm with the same base, then equate the inner arguments.

CRITICAL REQUIREMENT: Extraneous Solutions
Because the domain of $\log_b(g(x))$ is strictly $g(x) > 0$, any candidate solution that results in $\log_b(\text{zero})$ or $\log_b(\text{negative})$ in the ORIGINAL equation is extraneous and MUST be discarded!

Worked Example 4: Logarithmic Equation with Extraneous Root

Solve for $x$: $\log_2(x + 2) + \log_2(x - 2) = 5$

Solution:

1. Condense the left side using the Product Property $\log_2(u) + \log_2(v) = \log_2(u \cdot v)$: $\log_2\left((x + 2)(x - 2)\right) = 5 \implies \log_2(x^2 - 4) = 5$
2. Convert to exponential form ($u = b^c$): $x^2 - 4 = 2^5 \implies x^2 - 4 = 32$
3. Solve for $x$: $x^2 = 36 \implies x = \pm 6$
4. Check for Extraneous Solutions in original expression $\log_2(x + 2) + \log_2(x - 2) = 5$:
   • Check $x = 6$: $\log_2(6 + 2) + \log_2(6 - 2) = \log_2(8) + \log_2(4) = 3 + 2 = 5$. (Valid!)
   • Check $x = -6$: $\log_2(-6 + 2) = \log_2(-4)$, which is UNDEFINED. Thus $x = -6$ is extraneous.
5. The only valid solution is $x = 6$.

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3. Summary of Equation Types and Solving Strategies

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4. CLEP Exam Traps & Common Errors

• Trap 1: Forgetting to Check for Extraneous Log Roots
  Solving quadratic equations derived from logarithmic condensation often produces two algebraic roots. Failing to plug solutions back into original arguments leads to selecting invalid negative roots.
• Trap 2: Taking Logarithms Term-by-Term Incorrectly
  If given $e^x + e^{-x} = 4$, writing $\ln(e^x) + \ln(e^{-x}) = \ln(4)$ is completely false! You cannot take the log of individual terms in a sum; you must take the log of the ENTIRE side $\ln(e^x + e^{-x})$.
• Trap 3: Accepting Negative $u$-Values in Exponential Quadratic Substitutions
  When substituting $u = e^x$ in quadratic-form equations, if one root is $u = -3$, setting $e^x = -3$ yields no real solution because $e^x > 0$ for all real $x$.