Решите показательное уравнение! Помогите, оценка решается!!!!!!

[latex](3-2 \sqrt{2} )^{ x} + (3+2 \sqrt{2} )^{ x} = 34 |* (3+2 \sqrt{2} )^{ x}\\ (3-2 \sqrt{2} )^{ x}*(3+2 \sqrt{2} )^{ x} + (3+2 \sqrt{2} )^{ x}*(3+2 \sqrt{2} )^{ x}=34(3+2 \sqrt{2} )^{ x} \\ ((3-2 \sqrt{2} )(3+2 \sqrt{2} ))^{ x} + (3+2 \sqrt{2} )^{ 2x}=34(3+2 \sqrt{2} )^{ x} \\ [/latex] [latex]1 + (3+2 \sqrt{2} )^{ 2x}=34(3+2 \sqrt{2} )^{ x} \\ (3+2 \sqrt{2} )^{ 2x} - 34(3+2 \sqrt{2} )^{ x} + 1 = 0 \\ [/latex] Замена: [latex]t=(3+2 \sqrt{2})^{ x}\\ [/latex] [latex] t^{2} - 34t+ 1 = 0 \\ D=1156 - 4 = 1152, \sqrt{D} = \sqrt{1152}= \sqrt{16*36*2}= 24 \sqrt{2} \\ t_{1} = \frac{34+24 \sqrt{2} }{2} = 17+12 \sqrt{2} \\ t_{2} = \frac{34-24 \sqrt{2} }{2} = 17-12 \sqrt{2} \\ [/latex] Обратная замена: [latex](3+2 \sqrt{2})^{ x}=17+12 \sqrt{2} \\ ln((3+2 \sqrt{2})^{ x} )= ln(17+12 \sqrt{2} )\\ x *ln(3+2 \sqrt{2}) = ln(17+12 \sqrt{2} )\\ [/latex] [latex]x_{1}= \frac{ln(17+12 \sqrt{2} ) }{ln(3+2 \sqrt{2})} = log_{3+2 \sqrt{2}} (17+12 \sqrt{2})\\ [/latex] [latex](3+2 \sqrt{2})^{ x}=17-12 \sqrt{2} \\ ln((3+2 \sqrt{2})^{ x} )= ln(17-12 \sqrt{2} )\\ x *ln(3+2 \sqrt{2}) = ln(17-12 \sqrt{2} )\\x_{2}= \frac{ln(17-12 \sqrt{2} ) }{ln(3+2 \sqrt{2})} =log_{3+2 \sqrt{2}} (17-12 \sqrt{2})\\ [/latex]