PROSZE O POMOC
Rozwiąż równanie
a) [tex]\sqrt{3} ^{x^{2} -1} =9^{x+1}[/tex]
b)[tex]\sqrt{2} ^{x^{2} +4 } =4^{x+4}[/tex]
Rozwiąż równanie
a) [tex]\sqrt{3} ^{x^{2} -1} =9^{x+1}[/tex]
b)[tex]\sqrt{2} ^{x^{2} +4 } =4^{x+4}[/tex]
Odpowiedź:
rozwiązanie w załączniku
a)
[tex]\sqrt{3}^{x^{2}-1} = 9^{x+1}\\\\\sqrt{3}^{x^{2}-1} = (\sqrt{3}^{4})^{x+1}\\\\\sqrt{3}^{x^{2}-1} = \sqrt{3}{^{4(x+1)}}\\\\x^{2}-1 = 4(x+1)\\\\x^{2}-1 = 4x+4\\\\x^{2}-4x-1-4} = 0\\\\x^{2}-4x-5 = 0\\\\a = 1, \ b = -4, \ c = -5\\\\\Delta = b^{2}-4ac} = (-4)^{2}-4\cdot1\cdot(-5) = 16+20 = 36\\\\\sqrt{\Delta} = \sqrt{36} = 6\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{4-6}{2\cdot1} =\frac{-2}{2} = -1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{4+6}{2} = \frac{10}{2} = 5\\\\x\in\{-1,5\}[/tex]
b)
[tex]\sqrt{2}^{x^{2}+4} = 4^{x+4}\\\\\sqrt{2}^{x^{2}+4} = (\sqrt{2}^{4})^{x+4}\\\\\sqrt{2}^{x^{2}+4} = \sqrt{2}^{4(x+4)}\\\\x^{2}+4 = 4(x+4)\\\\x^{2}+4 = 4x+16\\\\x^{2}-4x+4-16 = 0\\\\x^{2}-4x-12 = 0\\\\a = 1, \ b = -4, \ c = -12\\\\\Delta} = b^{2}-4ac =(-4)^{2}-4\cdot1\cdot(-12) = 16+48 = 64\\\\\sqrt{\Delta} = \sqrt{64} = 8\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{4-8}{2\cdot1} = \frac{-4}{2} = -2\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{4+8}{2} = \frac{12}{2} = 6\\\\x \in\{-2,6\}[/tex]