Solving Logarithmic Equations: Concepts

The Search for 'x'

Solving a logarithmic equation is the process of finding the value of the unknown variable $x$ trapped inside a logarithm's argument or base. We use the product, quotient, and power rules of logarithms to condense multiple logarithmic terms into a single expression, and then convert it into its exponential form to solve.

The Golden Rule of Logarithmic Equations

You must always verify your solutions! Because the domain of $\log_b(a)$ is strictly positive ($a > 0$) and the base is positive and not equal to 1 ($b > 0, b \neq 1$), algebraic manipulations might produce **extraneous roots** (false solutions). If substituting an answer back results in the logarithm of a negative number or zero, that solution must be rejected.

Types of Logarithmic Equations

Logarithmic equations generally fall into one of the following categories, each requiring a specific algebraic strategy:

  • Type 1: Single Logarithmic Term
    Equations of the form $\log_b(f(x)) = c$.
    Strategy: Convert directly to exponential form: $f(x) = b^c$.
  • Type 2: Logarithms on Both Sides
    Equations of the form $\log_b(f(x)) = \log_b(g(x))$.
    Strategy: Equate the arguments directly: $f(x) = g(x)$ (ensuring $f(x)>0$ and $g(x)>0$).
  • Type 3: Equations Reducible to Quadratic Form
    Equations containing terms like $(\log_b x)^2$ and $\log_b x$.
    Strategy: Substitute $u = \log_b x$ to solve the quadratic equation $Au^2 + Bu + C = 0$, then solve for $x$ from $u$.
  • Type 4: Variable in the Base
    Equations of the form $\log_{g(x)}(f(x)) = c$.
    Strategy: Convert to exponential form: $f(x) = [g(x)]^c$, and verify $g(x) > 0$ and $g(x) \neq 1$.

Solved Examples (Examples 10 - 15)

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Example 10: Solve for $x$:   $\log_2(3x - 1) = 3$.
Show Solution
Convert the logarithmic statement into its equivalent exponential form:
$$3x - 1 = 2^3$$
$$3x - 1 = 8$$
$$3x = 9 \Rightarrow x = 3$$
Check: Substitute $x = 3$ back into the original expression:
$\log_2(3(3) - 1) = \log_2(8)$. Since $8 > 0$, the logarithm is defined.
Thus, the valid solution is **$x = 3$**.
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Example 11: Solve for $x$:   $\log_5(x) + \log_5(x - 4) = 1$.
Show Solution
Apply the Product Rule of logarithms:
$$\log_5[x(x - 4)] = 1$$
Convert to exponential form:
$$x(x - 4) = 5^1$$
$$x^2 - 4x - 5 = 0$$
$$(x - 5)(x + 1) = 0 \Rightarrow x = 5 \text{ or } x = -1$$
Check:
- If $x = 5$: the terms $\log_5(5)$ and $\log_5(1)$ are both defined ($5 > 0$, $1 > 0$). Valid!
- If $x = -1$: the term $\log_5(-1)$ is undefined. Thus, $x = -1$ is an **extraneous root** and must be rejected.

The only valid solution is **$x = 5$**.
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Example 12: Solve for $x$:   $\log_3(x^2 - 8x) = 2$.
Show Solution
Convert to exponential form:
$$x^2 - 8x = 3^2$$
$$x^2 - 8x - 9 = 0$$
$$(x - 9)(x + 1) = 0 \Rightarrow x = 9 \text{ or } x = -1$$
Check:
- If $x = 9$: $x^2 - 8x = 81 - 72 = 9 > 0$ (defined).
- If $x = -1$: $x^2 - 8x = 1 - 8(-1) = 9 > 0$ (defined).

Both solutions are valid. Therefore, **$x = 9, -1$**.
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Example 13: Solve for $x$:   $\log_2(3x - 4) = \log_2(5)$.
Show Solution
Since the bases are equal, equate the arguments directly:
$$3x - 4 = 5$$
$$3x = 9 \Rightarrow x = 3$$
Check: $3(3) - 4 = 5 > 0$ (defined).
The valid solution is **$x = 3$**.
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Example 14: Solve for $x$:   $(\log_3 x)^2 - 5(\log_3 x) + 6 = 0$.
Show Solution
Let $y = \log_3(x)$. Substitute $y$ into the equation:
$$y^2 - 5y + 6 = 0$$
$$(y - 3)(y - 2) = 0 \Rightarrow y = 3 \text{ or } y = 2$$
Now, convert back to solve for $x$:
- If $\log_3(x) = 3 \Rightarrow x = 3^3 = 27$.
- If $\log_3(x) = 2 \Rightarrow x = 3^2 = 9$.

Both $27 > 0$ and $9 > 0$, so both logarithms are defined.
The required solutions are **$x = 9, 27$**.
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Example 15: Solve for $x$:   $\log_b(\log_b x) = 1$  (where $b > 0, b \neq 1$).
Show Solution
Convert the outer logarithm to exponential form:
$$\log_b(x) = b^1 = b$$
Convert the remaining logarithm to exponential form:
$$\mathbf{x = b^b}$$
Enrichment Section

Graphs of Logarithmic & Exponential Functions

Logarithmic and exponential functions are mathematical inverses of each other. This inverse relationship can be clearly visualized on a coordinate graph:

y = x y = 2^x (0, 1) y = log_2(x) (1, 0) x y
  • The graph of $y = 2^x$ grows rapidly, passing through the point $(0, 1)$.
  • The graph of $y = \log_2(x)$ grows slowly, passing through the point $(1, 0)$.
  • Because they are inverses, their graphs are exact **mirror images** reflected across the diagonal identity line **$y = x$**.