Question

(+50) KuAT – Kuis Awal Tahun

Tentukan himpunan penyelesaian dari persamaan
[tex]\large\text{$\begin{aligned}\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}\:=\:x^{\sqrt{x}}\end{aligned}$}[/tex]

Answer 1

Penjelasan dengan langkah-langkah:

[tex]\large\text{$\begin{aligned}\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}\:=\:x^{\sqrt{x}}\end{aligned}$} \\ \sqrt[16]{ {x}^{15} } = {x}^{ {x}^{ \frac{1}{2} } } \\ {x}^{ \frac{15}{16} } = {x}^{ {x}^{ \frac{1}{2} } } \\ \frac{15}{16} = {x}^{ \frac{1}{2} } \\ ( \frac{15}{16} {)}^{2} = x \\ x = \frac{225}{256} [/tex]

15/16 Log x = √x Log x

x = 1

HP = {225/256 , 1}

Answer 2

Verified answer

Aljabar

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[tex]\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}\:=\:x^{\sqrt{x}} \\ \\ {x}^{ \frac{1}{2} ( \frac{1}{2} ( \frac{3}{2} \times \frac{1}{2} + 1) + 1)} \: = \: {x}^{ \sqrt{x} } \\ \\ {x}^{ \frac{15}{16} } \: = \: {x}^{ \sqrt{x} } \\ \\ [/tex]

x^(15/16) = x^√x

15/16 log x = √x log x

√x = 15/16 atau x = 1 (log 1 = 0)

x = (15/16)²

x = 225/256

HP = {225/256 , 1}