Given equation is \( \Large 4x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1} \) = \( \Large 2^{2x}+2^{2x-1} = 3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}} \) = \( \Large 2^{2x} \left(1+\frac{1}{2}\right)=3^{x-\frac{1}{2}} \left(3+1\right) \) = \( \Large 2^{2x}.\frac{3}{2}=3^{x-\frac{1}{2}}.4 => 2^{2x-3}=3^{x-\frac{3}{2}} \) taking log: \( \Large \left(2x-3\right)log2 = \left(x-\frac{3}{2}\right)log3 \) => \( \Large 2x log2 - 3 log2 = xlog3-\frac{3}{2}log3 \) => \( \Large x log 4 - x log 3 = 3 log^{2} - \frac{3}{2} log 3 \) = \( \Large xlog \left(\frac{4}{3}\right)= log 8 - log 3\sqrt{3} \) => \( \Large log \left(\frac{4}{3}\right)^{x} = log \frac{8}{3\sqrt{3}} \) => \( \Large \left(\frac{4}{3}\right)^{x} = \frac{8}{3\sqrt{3}} \) => \( \Large \left(\frac{4}{3}\right)^{x} = \left(\frac{4}{3}\right)^{\frac{3}{2}} \) \( \Large x = \frac{3}{2} \)
