Animated Solution for Mathematics - Basic Mathematics: Solve for $x$ the following equation : ${\log }_{(2x+3)}(6x^2+23x+21)=4-{\log }_{(3x+7)}(4x^2+12x+9)$.

Welcome, fellow traveler, to the beautiful world of logarithmic equations. When you first look at an equation like

it is natural to feel a surge of intimidation. It looks like a mess of quadratics and logs.

But here is the secret of the JEE Advanced topper: never start by calculating. Start by observing. The problem is not asking you to brute-force your way through; it is asking you to see the hidden symmetry.

The Hidden Factorization

Look closely at the arguments. We have $6x^2+23x+21$ and $4x^2+12x+9$. If you try to solve this using standard log expansion, you will be lost in a forest of complexity.

Instead, let us factorize. The first argument, $6x^2+23x+21$, splits beautifully into $(2x+3)(3x+7)$.

The second argument, $4x^2+12x+9$, is a perfect square: $(2x+3{)}^2$. Do you see it now? The bases of our logarithms are $(2x+3)$ and $(3x+7)$. The arguments are composed of these exact same factors. This is not a coincidence; it is a design.

The Logarithmic Dance

Now that we have revealed the structure, let us rewrite the equation:

Using the product rule, ${\log }_b(mn)={\log }_{b}m+{\log }_{b}n$, the left side becomes ${\log }_{(2x+3)}(2x+3)+{\log }_{(2x+3)}(3x+7)$. Since ${\log }_{b}b=1$, this simplifies to $1+{\log }_{(2x+3)}(3x+7)$.

On the right side, we use the power rule, ${\log }_b(m^n)=n{\log }_{b}m$, to bring that exponent of $2$ down, giving us $4-2{\log }_{(3x+7)}(2x+3)$. The equation is now:

The Substitution Strategy

We are staring at two log terms that are reciprocals of each other. Let $y={\log }_{(2x+3)}(3x+7)$. By the change of base property, ${\log }_{(3x+7)}(2x+3)=\frac{1}{y}$.

Our equation transforms into the elegant algebraic form:

Multiplying by $y$ gives us $y+y^2=4y-2$, which rearranges into the quadratic $y^2-3y+2=0$. Factoring this gives $(y-1)(y-2)=0$, leading to $y=1$ or $y=2$.

The Domain Minefield

This is where the battle is won or lost. We have two cases for $y$.

Case 1: ${\log }_{(2x+3)}(3x+7)=1$ leads to $3x+7=2x+3$, or $x=-4$. But wait! Check the base: $2(-4)+3=-5$. A negative base is undefined. Rejected.

Case 2: ${\log }_{(2x+3)}(3x+7)=2$ leads to $3x+7=(2x+3{)}^2$, which simplifies to $4x^2+9x+2=0$. Factoring this gives $(4x+1)(x+2)=0$, so $x=-\frac{1}{4}$ or $x=-2$.

Checking $x=-2$ in the base $3x+7$ gives $3(-2)+7=1$. A base of $1$ is undefined. Rejected. Finally, checking $x=-\frac{1}{4}$, we find valid bases.

The only survivor is $x=-\frac{1}{4}$. You have conquered the problem not just with algebra, but with mathematical rigor.