Comprehensive Solution to Logarithmic-Quadratic Equations for JEE 2026

Problem Definition and Statement

• The objective is to determine the number of solutions for a specific logarithmic-quadratic equation geared toward JEE 2026 preparation.
• The provided equation is:     $\log_{(x-3)}(2x^2 - 7x + 3) = 2 \log_{(2x-1)}(x-3)$

Fundamental Logarithmic Constraints and Domain Verification

Before solving the algebraic structure of the equation, the domain of the variable $x$ must be strictly defined based on the properties of logarithms. For any expression $\log_b(a)$, the following conditions must be met:

• The base $b$ must be greater than zero ($b > 0$).
• The base $b$ must not equal one ($b \neq 1$).
• The argument $a$ must be greater than zero ($a > 0$).

Condition Set 1: Based on $\log_{(x-3)}(2x^2 - 7x + 3)$

• Base Constraint: $x - 3 > 0 \implies x > 3$
• Base Unity Constraint: $x - 3 \neq 1 \implies x \neq 4$
• Argument Constraint: $2x^2 - 7x + 3 > 0$
  • Factoring the quadratic: $(2x - 1)(x - 3) > 0$
  • The roots of the quadratic are $x = \frac{1}{2}$ and $x = 3$.
  • The inequality holds for $x < \frac{1}{2}$ or $x > 3$.

Condition Set 2: Based on $2 \log_{(2x-1)}(x-3)$

• Base Constraint: $2x - 1 > 0 \implies x > \frac{1}{2}$
• Base Unity Constraint: $2x - 1 \neq 1 \implies 2x \neq 2 \implies x \neq 1$
• Argument Constraint: $x - 3 > 0 \implies x > 3$

Combined Global Domain

• By intersecting all constraints ($x > 3$, $x \neq 4$, $x < \frac{1}{2} \cup x > 3$, $x > \frac{1}{2}$, and $x \neq 1$), the permissible domain for $x$ is:     $x \in (3, 4) \cup (4, \infty)$

Algebraic Transformation of the Equation

• The equation includes a quadratic term in the first argument: $\log_{(x-3)}((2x - 1)(x - 3)) = 2 \log_{(2x-1)}(x-3)$.
• Utilize the logarithmic identity $\log_b(m \times n) = \log_b(m) + \log_b(n)$.     $\log_{(x-3)}(2x - 1) + \log_{(x-3)}(x - 3) = 2 \log_{(2x-1)}(x - 3)$
• Since $\log_b(b) = 1$, the equation simplifies to:     $\log_{(x-3)}(2x - 1) + 1 = 2 \log_{(2x-1)}(x - 3)$
• Apply the change of base law or the reciprocal property $\log_a(b) = \frac{1}{\log_b(a)}$.     $\log_{(x-3)}(2x - 1) + 1 = \frac{2}{\log_{(x-3)}(2x - 1)}$

Method of Substitution and Solving for $t$

• Let $t = \log_{(x-3)}(2x - 1)$.
• Substitute $t$ into the simplified equation:     $t + 1 = \frac{2}{t}$
• Multiply the entire equation by $t$ to clear the fraction (assuming $t \neq 0$):     $t^2 + t = 2 \implies t^2 + t - 2 = 0$
• Factor the quadratic in terms of $t$:     $(t + 2)(t - 1) = 0$
• This yields two potential values for $t$:
  1. $t = 1$
  2. $t = -2$

Back-Substitution and solving for $x$

Case 1: $t = 1$

• $\log_{(x-3)}(2x - 1) = 1$
• By definition of logarithms: $2x - 1 = (x - 3)^1$
• Solve for $x$:     $2x - 1 = x - 3$$x = -2$
• Verification: Check against global domain ($x > 3$). Since $-2$ is not greater than $3$, this solution is rejected.

Case 2: $t = -2$

• $\log_{(x-3)}(2x - 1) = -2$
• By definition: $(x - 3)^{-2} = 2x - 1$
• Rewrite using the reciprocal: $\frac{1}{(x - 3)^2} = 2x - 1$
• Cross-multiply to form a cubic equation:     $1 = (2x - 1)(x - 3)^2$$1 = (2x - 1)(x^2 - 6x + 9)$$1 = 2x^3 - 12x^2 + 18x - x^2 + 6x - 9$$2x^3 - 13x^2 + 24x - 10 = 0$

Solving the Cubic Equation

• Test for rational roots using the Factor Theorem. Checking $x = 2.5$ or $x = \frac{5}{2}$.     $2(\frac{5}{2})^3 - 13(\frac{5}{2})^2 + 24(\frac{5}{2}) - 10$$\frac{125}{4} - \frac{325}{4} + 60 - 10 = -50 + 50 = 0$
• Thus, $(2x - 5)$ is a factor. Dividing the cubic by $(2x - 5)$ results in a quadratic factor:     $(2x - 5)(x^2 - 4x + 2) = 0$
• Finding the roots of $x^2 - 4x + 2 = 0$ using the quadratic formula:     $x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$$x = \frac{4 \pm \sqrt{16 - 8}}{2}$$x = \frac{4 \pm 2\sqrt{2}}{2}$$x = 2 \pm \sqrt{2}$

Final Solution Verification

We have three candidates from Case 2 to check against the domain constraint $x \in (3, 4) \cup (4, \infty)$.

1. Candidate $x = 2.5$: Since $2.5 < 3$, this solution is rejected.
2. Candidate $x = 2 - \sqrt{2}$: Since $2 - 1.414 = 0.586 < 3$, this solution is rejected.
3. Candidate $x = 2 + \sqrt{2}$: Since $2 + 1.414 = 3.414$, this value falls between $3$ and $4$. It satisfy the domain constraint and the base unity constraint ($3.414 \neq 4$).

Result Summary

• There is exactly one solution to the equation.
• The valid solution is $x = 2 + \sqrt{2}$.