Question

Resolva com IR, a equação exponencial: (\sqrt{2})^{3x-1}=(\sqrt[3]{16})^{2x-1}

Answer 1

$(\sqrt{2})^{3x-1}=(\sqrt[3]{16})^{2x-1}\\ \\ (2^{\frac{1}{2}})^{3x-1}=( \sqrt[3]{2^4})^{2x-1}\\ \\ (2^{\frac{1}{2}})^{3x-1}=(2^{\frac{4}{3}} )^{2x-1}\\ \\ 2^{\frac{3x-1}{2}}=2^{\frac{8x-4}{3}}\\ \\ \frac{3x-1}{2}=\frac{8x-4}{3}\\ \\ 2(8x-4)=3(3x-1)\\ \\ 16x-8=9x-3\\ \\ 16x-9x=8-3\\ \\ 7x=5\\ \\ x=\frac{5}{7}$