Usando as propriedades,calcule:
log₂ 16√8 =
log₃ 81√3 =
∛3
Alguém me ajude,por favor!
Soluções para a tarefa
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Lembrando:
![\sqrt[n]{a^{x}}=a^{x/n}](https://tex.z-dn.net/?f=+%5Csqrt%5Bn%5D%7Ba%5E%7Bx%7D%7D%3Da%5E%7Bx%2Fn%7D+ "\sqrt[n]{a^{x}}=a^{x/n}")
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![log_{2}(16\sqrt{8})=log_{2}(16\sqrt[2]{2^{3}}) \\ log_{2}(16\sqrt{8})=log_{2}(2^{4}*2^{3/2}) \\ log_{2}(16\sqrt{8})=log_{2}(2^{4+[3/2]})\\log_{2}(16\sqrt{8})=log_{2}(2^{11/2})\\log_{2}(16\sqrt{8}=(11/2)*log_{2}(2)\\log_{2}(16\sqrt{8})=(11/2)*1\\log_{2}(16\sqrt{8})=11/2](https://tex.z-dn.net/?f=log_%7B2%7D%2816%5Csqrt%7B8%7D%29%3Dlog_%7B2%7D%2816%5Csqrt%5B2%5D%7B2%5E%7B3%7D%7D%29+%5C%5C+log_%7B2%7D%2816%5Csqrt%7B8%7D%29%3Dlog_%7B2%7D%282%5E%7B4%7D%2A2%5E%7B3%2F2%7D%29+%5C%5C+log_%7B2%7D%2816%5Csqrt%7B8%7D%29%3Dlog_%7B2%7D%282%5E%7B4%2B%5B3%2F2%5D%7D%29%5C%5Clog_%7B2%7D%2816%5Csqrt%7B8%7D%29%3Dlog_%7B2%7D%282%5E%7B11%2F2%7D%29%5C%5Clog_%7B2%7D%2816%5Csqrt%7B8%7D%3D%2811%2F2%29%2Alog_%7B2%7D%282%29%5C%5Clog_%7B2%7D%2816%5Csqrt%7B8%7D%29%3D%2811%2F2%29%2A1%5C%5Clog_%7B2%7D%2816%5Csqrt%7B8%7D%29%3D11%2F2 "log_{2}(16\sqrt{8})=log_{2}(16\sqrt[2]{2^{3}}) \\ log_{2}(16\sqrt{8})=log_{2}(2^{4}*2^{3/2}) \\ log_{2}(16\sqrt{8})=log_{2}(2^{4+[3/2]})\\log_{2}(16\sqrt{8})=log_{2}(2^{11/2})\\log_{2}(16\sqrt{8}=(11/2)*log_{2}(2)\\log_{2}(16\sqrt{8})=(11/2)*1\\log_{2}(16\sqrt{8})=11/2")
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![log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{4}*3^{1/2}/3^{1/3})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{4+[1/2]-[1/3]})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{[24+3-2]/6})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{25/6})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=(25/6)*log_{3}(3)\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=(25/6)*1\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=25/6](https://tex.z-dn.net/?f=log_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3Dlog_%7B3%7D%283%5E%7B4%7D%2A3%5E%7B1%2F2%7D%2F3%5E%7B1%2F3%7D%29%5C%5Clog_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3Dlog_%7B3%7D%283%5E%7B4%2B%5B1%2F2%5D-%5B1%2F3%5D%7D%29%5C%5Clog_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3Dlog_%7B3%7D%283%5E%7B%5B24%2B3-2%5D%2F6%7D%29%5C%5Clog_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3Dlog_%7B3%7D%283%5E%7B25%2F6%7D%29%5C%5Clog_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3D%2825%2F6%29%2Alog_%7B3%7D%283%29%5C%5Clog_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3D%2825%2F6%29%2A1%5C%5Clog_%7B3%7D%2881%5Csqrt%7B3%7D%2F%5Csqrt%5B3%5D%7B3%7D%29%3D25%2F6 "log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{4}*3^{1/2}/3^{1/3})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{4+[1/2]-[1/3]})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{[24+3-2]/6})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=log_{3}(3^{25/6})\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=(25/6)*log_{3}(3)\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=(25/6)*1\\log_{3}(81\sqrt{3}/\sqrt[3]{3})=25/6")
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flavyane1:
Obrigado,mas não entendi por que na parte 4+3/2 deu 11?
4 + (3/2) = (8 / 2) + (3 / 2)
Quando os denominadores são iguais, os conservamos e somamos os numeradores: (8 + 3) / 2 = 11 / 2
Muito obrigado!
Nada ;D
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