[tex]\left(3\big2^{-5}\cdot\dfrac1{256}\right)^{2\frac23}:\left(\sqrt[3]{128}:16\right) = \left(\left(2^5\right)^{-5}\cdot\dfrac1{2^8}\right)^{2\frac23}:\left(\sqrt[3]{2^7}:\big2^4\right)= \\\\= \left(\big2^{-25}\cdot\big2^{-8}}\right)^{2\frac23}:\left(\big2^\frac73}:\big2^\frac{12}3}\right)= \left(\big2^{-25+(-8)}}\right)^{2\frac23}:\left(\big2^{\frac73-\frac{12}3}\right)=[/tex]
[tex]= \left(\big2^{-33}}\right)^{\frac83}:\big2^{-\frac{5}3}= \big2^{-33\cdot\frac83}:\big2^{-\frac{5}3}=\big2^{-88}:\big2^{-1\frac{2}3} =\big2^{-88-(-1\frac{2}3)}=\big2^{-86\frac{1}3}[/tex]
[tex]\big3^{2x-6}\cdot24\big3^{x-2}=\dfrac{3\sqrt3}{2\big7^{x-1}}\qquad\ /\cdot2\big7^{x-1} \\\\\big3^{2x-6}\cdot24\big3^{x-2}\cdot2\big7^{x-1}=3\sqrt3 \\\\ \big3^{2x-6}\cdot\big(3^5\big)^{x-2}\cdot\big(3^3\big)^{x-1}= \big3^1\cdot\big3^\frac12\\\\\big3^{2x-6}\cdot\big3^{5\cdot(x-2)}\cdot \big 3^{3\cdot(x-1)} = \big3^{1+\frac12}\\\\\big3^{2x-6+5\cdot(x-2)+3\cdot(x-1)} = \big3^{1\frac12}\\\\ 2x-6+5\cdot(x-2)+3\cdot(x-1)=1\frac12\\\\ 2x-6+5x-10+3x-3= 1\frac12\qquad/+19\\\\ 10x=20\frac12\qquad/:10\\\\x=2\frac1{20}[/tex]
