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Question:

If $9^{x-\frac{1}{2}}$ - 2^{2x - 2} = 4^x - 3^{2x - 3}, then x is

Explanation:

$9^{\mathrm{x}-\frac{1}{2}}$ - 2^{2x - 2} = 4^x - 3^{2x - 3}   ...(1)

Now, $9^{\mathrm{x}-\frac{1}{2}}$ = ${\left(3^2\right)}^{x-\frac{1}{2}}$ = 3^{2x - 1}

Also 4^x = (2²)^x = 2^2x

From (1), we get

3^{2x - 1} - 2^{2x - 2} = 2^2x - 3^{2x - 3}

3^{2x - 1} + 3^{2x - 3} = 2^2x + 2^{2x - 2}

3^{2x - 3} (3² + 1) = 2^{2x - 2} (2² + 1)

$\frac{3^{2\mathrm{x}-3}}{2^{2\mathrm{x}-2}}=\frac{5}{10} \mathrm{or} \frac{1}{2}$

Now this can be written as

$\frac{3^{2x-3}}{2^{2x-2}}=\frac{3^0}{2^1}$

Now since the bases of the numerator and the dominator are equal on LHS and RHS, the powers should also be equal 

⇒ 3^2x-3 = 3⁰ and 2^{2x - 2} = 2¹

In both cases x = 3/2

∴ x = $\frac{3}{2}$

Hence, option (a).