JEE Main Exponential Equation Trick | Solve in Under 2 Minutes

 

❓ Question

Find the unique value of $x$ satisfying

$$4^x-3^{x+\frac12}=3^{x+\frac12}-2^{\,2x-1}.$$

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✍️ Solution

Given,

$$4^x-3^{x+\frac12}=3^{x+\frac12}-2^{\,2x-1}$$

Since,

$$4^x=2^{2x},\qquad 3^{x+\frac12}=3^x\sqrt3,\qquad 2^{2x-1}=\frac{2^{2x}}2$$

Substitute:

$$2^{2x}-3^x\sqrt3 = 3^x\sqrt3-\frac{2^{2x}}2$$

Rearrange,

$$2^{2x}+\frac{2^{2x}}2 = 3^x\sqrt3+3^x\sqrt3$$
$$2^{2x}\left(1+\frac12\right) = 2\sqrt3\,3^x$$
$$2^{2x}\cdot\frac32 = 2\sqrt3\,3^x$$
$$2^{2x} = \frac{4\sqrt3}{3}\,3^x$$

Hence,

$$\left(\frac43\right)^x = \frac{4\sqrt3}{3} = \frac{4}{\sqrt3} = \frac{2^2}{3^{1/2}}$$

Also,

$$\left(\frac43\right)^x = \left(\frac{2^2}{3}\right)^x = \frac{2^{2x}}{3^x}$$

Comparing powers,

$$\frac{2^{2x}}{3^x} = \frac{2^2}{3^{1/2}}$$

Therefore,

$$2x=2,\qquad x=\frac12.$$

(And $x=\frac12$ also satisfies the power of $3$: $x=\frac12$.)

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✅ Final Answer

$$\boxed{x=\frac12}$$