Question

Resolva a equação log2(7^4x-2 + 7) =3 +log2 (7 ^2x-1)

Answer 1

$log_{2} (7^{4x-2}+7)=3+log_{2}(7^{2x-1})\\log_{2} (7^{4x-2}+7)-log_{2}(7^{2x-1})=3\\log_{2}(\frac{7^{4x-2}+7}{7^{2x-1}})=3\\\frac{7^{4x-2}+7}{7^{2x-1}} =2^3\\\frac{7^{4x}.7^{-2}+7}{7^{2x}.7^{-1}}=8\\ \frac{(7^{2x})^2.\frac{1}{7^2}+7 }{7^{2x}.\frac{1}{7} } =8\\\frac{(7^{2x})^2.\frac{1}{49}+7 }{7^{2x}.\frac{1}{7} } =8\\\frac{t^2.\frac{1}{49}+7 }{t.\frac{1}{7} } \\t=7\\t=49\\\\7^{2x}=7\\7^{2x}=49\\\\x=\frac{1}{2}\\ x=1\\\\x_1=\frac{1}{2}\\ x_2=1$