Logarithmic and Exponential Equations
Key Definitions
- Logarithmic Equation: An equation that contains a logarithm of a variable expression. For example, $\log_a (x) = b$ or $\log_a (x) + \log_a (x-1) = 2$.
- Exponential Equation: An equation in which the variable appears in an exponent. For example, $2^x = 8$ or $3^{2x-1} = 27$.
- Solution Set of Logarithmic Equations: The set of all values of the variable that satisfy the equation and also satisfy the domain conditions (arguments of all logarithms must be positive).
- Principal Method for Solving: Convert logarithmic equations to exponential form, or convert exponential equations to logarithmic form. For equations with multiple logarithms, combine them using logarithmic laws.
- Extraneous Solutions: Solutions obtained algebraically that do not satisfy the domain conditions of the original logarithmic equation. These must be rejected.
Theory and Concepts
Solving Logarithmic Equations
Method 1: Converting to Exponential Form
For an equation of the form $\log_a f(x) = b$ (where $a > 0$, $a \neq 1$):
$$ \log_a f(x) = b \iff f(x) = a^b $$Always check that $f(x) > 0$.
Method 2: Using the One-to-One Property
If $\log_a f(x) = \log_a g(x)$, then $f(x) = g(x)$ (provided $f(x) > 0$ and $g(x) > 0$).
Method 3: Combining Logarithms
For equations with sums/differences of logarithms:
- Combine logs using product/quotient rules
- Convert to exponential form
- Solve the resulting algebraic equation
- Verify domain conditions
Solving Exponential Equations
Method 1: Same Base Method
If $a^{f(x)} = a^{g(x)}$ and $a > 0$, $a \neq 1$, then $f(x) = g(x)$.
Method 2: Taking Logarithms
For equations of the form $a^{f(x)} = b$ (where $b > 0$):
$$ a^{f(x)} = b \implies f(x) = \log_a b $$Method 3: Using Natural Logarithms
For equations like $p^x = q$:
$$ \ln(p^x) = \ln q \implies x \ln p = \ln q \implies x = \frac{\ln q}{\ln p} $$Important Strategies
Domain Conditions for Logarithmic Equations
For any logarithmic equation, the following must always hold:
- Base $a > 0$ and $a \neq 1$
- Argument of each logarithm $> 0$
- For $\log_a f(x)$, we require $f(x) > 0$
Solved Examples
Solved Example 4.1
Solve: $\log_2 (x^2 – 5x + 6) = 1$.
Solution:
Convert to exponential form:
$$ x^2 – 5x + 6 = 2^1 = 2 $$ $$ x^2 – 5x + 4 = 0 $$ $$ (x – 1)(x – 4) = 0 \implies x = 1 \text{ or } x = 4 $$Domain condition: $x^2 – 5x + 6 > 0$
$$ (x – 2)(x – 3) > 0 \implies x < 2 \text{ or } x > 3 $$Both $x = 1$ and $x = 4$ satisfy the domain condition. Therefore, the solution set is $\{1, 4\}$.
Solved Example 4.2
Solve: $\log_3 (x + 2) + \log_3 (x – 1) = 2$.
Solution:
Using product rule:
$$ \log_3 [(x + 2)(x – 1)] = 2 $$Convert to exponential form:
$$ (x + 2)(x – 1) = 3^2 = 9 $$ $$ x^2 + x – 2 = 9 $$ $$ x^2 + x – 11 = 0 $$ $$ x = \frac{-1 \pm \sqrt{1 + 44}}{2} = \frac{-1 \pm \sqrt{45}}{2} = \frac{-1 \pm 3\sqrt{5}}{2} $$Domain conditions: $x + 2 > 0 \implies x > -2$ and $x – 1 > 0 \implies x > 1$. So $x > 1$.
Now $\frac{-1 + 3\sqrt{5}}{2} \approx 2.854 > 1$, valid.
$\frac{-1 – 3\sqrt{5}}{2} \approx -3.854 < 1$, invalid.
Hence, $x = \frac{-1 + 3\sqrt{5}}{2}$.
Solved Example 4.3
Solve: $\log_2 (x + 1) – \log_2 (x – 1) = \log_2 3$.
Solution:
Using quotient rule:
$$ \log_2 \left( \frac{x + 1}{x – 1} \right) = \log_2 3 $$Using one-to-one property:
$$ \frac{x + 1}{x – 1} = 3 $$ $$ x + 1 = 3x – 3 $$ $$ 4 = 2x \implies x = 2 $$Domain conditions: $x + 1 > 0 \implies x > -1$ and $x – 1 > 0 \implies x > 1$. So $x > 1$.
$x = 2$ satisfies $x > 1$. Therefore, $x = 2$ is the solution.
Solved Example 4.4
Solve: $2^{2x} – 5 \cdot 2^x + 4 = 0$.
Solution:
Let $y = 2^x$. Then the equation becomes:
$$ y^2 – 5y + 4 = 0 $$ $$ (y – 1)(y – 4) = 0 \implies y = 1 \text{ or } y = 4 $$Since $y = 2^x > 0$, both are valid.
$$ 2^x = 1 \implies x = 0 $$ $$ 2^x = 4 \implies x = 2 $$Therefore, $x = 0$ or $x = 2$.
Solved Example 4.5
Solve: $3^{x+2} = 5^{x-1}$.
Solution:
Take natural logarithm on both sides:
$$ \ln(3^{x+2}) = \ln(5^{x-1}) $$ $$ (x + 2)\ln 3 = (x – 1)\ln 5 $$ $$ x\ln 3 + 2\ln 3 = x\ln 5 – \ln 5 $$ $$ x\ln 3 – x\ln 5 = -\ln 5 – 2\ln 3 $$ $$ x(\ln 3 – \ln 5) = -(\ln 5 + 2\ln 3) $$ $$ x = \frac{-(\ln 5 + 2\ln 3)}{\ln 3 – \ln 5} = \frac{\ln 5 + 2\ln 3}{\ln 5 – \ln 3} = \frac{\ln(5 \times 3^2)}{\ln(5/3)} = \frac{\ln 45}{\ln(5/3)} $$Solved Example 4.6
Solve: $\log_2 x + \log_4 x + \log_8 x = \frac{11}{6}$.
Solution:
Convert all logs to base 2:
$$ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} $$ $$ \log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{\log_2 x}{3} $$Let $y = \log_2 x$. Then:
$$ y + \frac{y}{2} + \frac{y}{3} = \frac{11}{6} $$ $$ \frac{6y + 3y + 2y}{6} = \frac{11}{6} \implies \frac{11y}{6} = \frac{11}{6} \implies y = 1 $$Therefore, $\log_2 x = 1 \implies x = 2$.
Solved Example 4.7
Solve: $2^{x} + 2^{x+2} = 20$.
Solution:
$$ 2^x + 2^x \cdot 2^2 = 20 $$ $$ 2^x + 4 \cdot 2^x = 20 $$ $$ 5 \cdot 2^x = 20 $$ $$ 2^x = 4 = 2^2 \implies x = 2 $$Solved Example 4.8
Solve: $\log_x 2 \cdot \log_{x/16} 2 = \log_{x/64} 2$.
Solution:
Using change of base to base 2:
$$ \log_x 2 = \frac{1}{\log_2 x}, \quad \log_{x/16} 2 = \frac{1}{\log_2(x/16)} = \frac{1}{\log_2 x – 4} $$ $$ \log_{x/64} 2 = \frac{1}{\log_2(x/64)} = \frac{1}{\log_2 x – 6} $$Let $t = \log_2 x$. Then:
$$ \frac{1}{t} \cdot \frac{1}{t-4} = \frac{1}{t-6} \implies \frac{1}{t(t-4)} = \frac{1}{t-6} $$ $$ t-6 = t(t-4) \implies t-6 = t^2 – 4t $$ $$ 0 = t^2 – 5t + 6 \implies (t-2)(t-3) = 0 \implies t = 2 \text{ or } t = 3 $$Domain conditions: $x > 0$, $x \neq 1$, $x/16 > 0$, $x/64 > 0$, and $x/16 \neq 1$, $x/64 \neq 1$.
$$ t = 2 \implies x = 4 $$ $$ t = 3 \implies x = 8 $$Both satisfy domain conditions. Therefore, $x = 4$ or $x = 8$.
Solved Example 4.9
Solve: $\log_5 (3x + 2) = \log_5 (2x + 5)$.
Solution:
Using one-to-one property:
$$ 3x + 2 = 2x + 5 \implies x = 3 $$Domain conditions: $3x + 2 > 0 \implies x > -\frac{2}{3}$ and $2x + 5 > 0 \implies x > -\frac{5}{2}$. $x = 3$ satisfies both. Therefore, $x = 3$.
Solved Example 4.10
Solve: $e^{2x} – 3e^x + 2 = 0$.
Solution:
Let $y = e^x > 0$. Then:
$$ y^2 – 3y + 2 = 0 $$ $$ (y – 1)(y – 2) = 0 \implies y = 1 \text{ or } y = 2 $$Since $y = e^x$:
$$ e^x = 1 \implies x = 0 $$ $$ e^x = 2 \implies x = \ln 2 $$Therefore, $x = 0$ or $x = \ln 2$.
Concept Application Exercise 4.1
- Solve: $\log_3 (x^2 – 4x + 3) = 1$.
- Solve: $\log_2 (x – 1) + \log_2 (x – 2) = 1$.
- Solve: $\log_5 (x + 2) – \log_5 (x – 2) = \log_5 2$.
- Solve: $4^x – 3 \cdot 2^x + 2 = 0$.
- Solve: $2^{x+1} = 3^{x-2}$.
- Solve: $\log_2 x + \log_4 x + \log_{16} x = \frac{21}{4}$.
- Solve: $3^{2x} – 4 \cdot 3^x + 3 = 0$.
- Solve: $\log_x 3 \cdot \log_{x/9} 3 = \log_{x/27} 3$.
- Solve: $\log_3 (2x + 1) = \log_3 (x^2 – 2x + 5)$.
- Solve: $e^{2x} – 5e^x + 6 = 0$.