Animated Solution for Mathematics - Basic Mathematics: Solve the following equation for $x:2{\log }_{x}a+{\log }_{ax}a+3{\log }_{a^{2}x}a=0,a>0,a\ne 0$

Welcome, fellow traveler on the JEE journey. Today, we are going to tackle a problem that looks intimidating at first glance but reveals a beautiful, symmetrical structure once we peel back the layers.

We are looking at the equation:

The first thing that should catch your eye is the variable $x$ sitting in the base of every single term. In the world of logarithms, having a variable in the base is like trying to build a house on shifting sand. It is unstable, and it makes direct manipulation nearly impossible.

But look closer at the arguments—they are all just $a$. This is our golden ticket.

To solve this, we need to get $x$ out of the basement. We invoke the powerful base-change property:

By taking the reciprocal, we can flip the base and the argument. This transforms our equation into:

Suddenly, the variable $x$ is in the argument, and the base is a constant $a$. The problem has shifted from a chaotic mess to a structured, solvable form. This is the hallmark of a great JEE problem—it tests your ability to recognize when to change your perspective.

Now, we have denominators like ${\log }_a(ax)$ and ${\log }_a(a^{2}x)$. We need to break these apart.

Using the product property, ${\log }_a(mn)={\log }_{a}m+{\log }_{a}n$, we can expand these. ${\log }_a(ax)$ becomes ${\log }_{a}a+{\log }_{a}x$, which simplifies to $1+{\log }_{a}x$.

Similarly, ${\log }_a(a^{2}x)$ becomes ${\log }_a{a}^2+{\log }_{a}x$, which is $2+{\log }_{a}x$. Our equation now looks like this:

Writing ${\log }_{a}x$ repeatedly is not just tedious; it is an invitation for a silly mistake. Let's introduce a dummy variable, $t={\log }_{a}x$.

The equation becomes a clean, rational algebraic equation:

To clear the fractions, we multiply the entire equation by the lowest common multiple, $t(1+t)(2+t)$. This gives us:

Expanding this carefully, we get $2(t^2+3t+2)+(t^2+2t)+(3t^2+3t)=0$. Combining like terms, we arrive at the quadratic:

We solve this quadratic by splitting the middle term. We need two numbers that multiply to $6\times 4=24$ and add to $11$. Those numbers are $8$ and $3$.

So, $6t^2+8t+3t+4=0$, which factors into $(2t+1)(3t+4)=0$. This gives us two values for $t$:

Finally, we substitute back $t={\log }_{a}x$. This means $x=a^t$.

Thus, our solutions are $x=a^{-1/2}$ and $x=a^{-4/3}$. Take a moment to appreciate the elegance of this result. We started with a complex logarithmic equation and, through systematic simplification, reduced it to a simple quadratic.