Animated Solution for Mathematics - Basic Mathematics: Solve for $x:4^x-3^{x-1/2}=3^{x+1/2}-2^{2x-1}$

Imagine you are standing before a complex exponential equation: $4^x-3^{x-1/2}=3^{x+1/2}-2^{2x-1}$. At first glance, it looks like a chaotic jumble of bases and fractional exponents.

In the world of JEE Advanced, chaos is just order waiting to be discovered. The secret to mastering these problems is to stop seeing a single equation and start seeing 'families'.

We have a base-2 family ($4^x$ and $2^{2x-1}$) and a base-3 family ($3^{x+1/2}$ and $3^{x-1/2}$). By moving the negative terms to their respective sides, we transform the equation into:

Now, we must confront the fractional exponents. Recall the fundamental law of indices: $a^{m-n}=\frac{a^m}{a^n}$.

For the base-2 family, the term $2^{2x-1}$ is simply $\frac{2^{2x}}{2}$. Since $2^{2x}$ is $(2^2{)}^x$, which is $4^x$, our term becomes $\frac{4^x}{2}$.

Similarly, for the base-3 family, $3^{x+1/2}$ becomes $3^x\cdot \sqrt{3}$, and $3^{x-1/2}$ becomes $\frac{3^x}{\sqrt{3}}$. Substituting these back, our equation becomes:

We are now at the most satisfying part of the process: the extraction. Notice that both sides now have a common exponential term.

On the left, we factor out $4^x$ to get $4^x(1+\frac{1}{2})$, which simplifies to $4^x(\frac{3}{2})$. On the right, we factor out $3^x$ to get $3^x(\sqrt{3}+\frac{1}{\sqrt{3}})$.

Simplifying the bracket on the right, we find $\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{3+1}{\sqrt{3}}=\frac{4}{\sqrt{3}}$. Now, we have:

To solve for $x$, we isolate the variable by dividing both sides by $3^x$ and multiplying by the reciprocal of the constant on the left. This yields:

The final hurdle is to express the right side as a power of $\frac{4}{3}$. Since $\frac{8}{3\sqrt{3}}={\left(\frac{2}{\sqrt{3}}\right)}^3$ and $\frac{2}{\sqrt{3}}={\left(\frac{4}{3}\right)}^{1/2}$, the right side is simply:

Equating the exponents, we arrive at the final result: