Logaritmo

por KOSHAI Sáb 03 Abr 2021, 19:06

Resolva a equação:

[latex]\sqrt{1+log_{2}x}+\sqrt{4\cdot log_{4}x-2}=4[/latex]

Gabarito:

S= {8}:

por **Eduardo Rabelo** Sáb 03 Abr 2021, 21:13

$Logaritmo Gif.latex?%5Cdpi%7B120%7D%20%5Cfn_cm%20%5C%5C%5Csqrt%7B%5Clog_%7B2%7D2x%7D%20+%5Csqrt%7B2%5Cleft%28%5Clog_%7B2%7D%5Cfrac%7Bx%7D%7B2%7D%20%5Cright%20%29%7D%3D4%5Ctherefore%20%5C%5C%5C%5C%5Ctherefore%20%5Clog_%7B2%7D2x+%5Clog_%7B2%7D%5Cfrac%7Bx%5E%7B2%7D%7D%7B4%7D+2%5Csqrt%7B%5Clog_%7B2%7D2x%5Ccdot%20%5Clog_%7B2%7D%5Cfrac%7Bx%5E%7B2%7D%7D%7B4%7D%7D%3D16%5Ctherefore%20%5C%5C%5C%5C%5Ctherefore%20%5Cleft%2817-%5Clog_%7B2%7Dx%5E%7B3%7D%20%5Cright%20%29%5E%7B2%7D%3D8%5Cleft%281+%5Clog_%7B2%7Dx%20%5Cright%20%29%5Cleft%28%5Clog_%7B2%7Dx-1%20%5Cright%20%29%3D8%28%5Clog%5E%7B2%7D_%7B2%7Dx-1%29%5Ctherefore%20%5C%5C%5C%5C%5Ctherefore%20289+%5Clog_%7B2%7D%5E%7B2%7D%20x%5E%7B3%7D-34%5Clog_%7B2%7Dx%5E%7B3%7D%3D8%5Clog%5E%7B2%7D_%7B2%7Dx-8%5Ctherefore%20%5C%5C%5C%5C%5Ctherefore%20%5Clog%5E%7B2%7D_%7B2%7Dx-102%5Clog_%7B2%7Dx+297%3D0%5Ctherefore%20%5C%5C%5C%5C%5Ctherefore%20%5Clog_%7B2%7Dx%3D99%5Ctherefore%20%5Cunderset%7BN%5Ctilde%7Ba%7Do%5C%3Bserve$

Se fosse numa prova eu acho que não faria essa conta, pois é uma raiz um pouco trivial.

por Fibonacci13 Sáb 03 Abr 2021, 21:20

Outra maneira:

[latex]\sqrt{1+\log _2\left(x\right)}+\sqrt{4\log _4\left(x\right)-2}=4[/latex]

[latex]\sqrt{1+2\log _4\left(x\right)}+\sqrt{4\log _4\left(x\right)-2}=4[/latex]

[latex]\sqrt{1+2k}+\sqrt{4k-2}=4[/latex]

[latex]k=\frac{3}{2}[/latex]

Substituindo o K:

[latex]x = 8[/latex]

por Conteúdo patrocinado

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