Question

Soal persamaan eksponen yang menantang. Berapa nilai x yang memenuhi?
[tex]\large 2^{x^4-10x^2+8}=\sqrt{\frac{x}{x^2+3}}[/tex]

Answer 1

Jawab:

Penjelasan dengan langkah-langkah:

[tex]2^{x^{4} - 10x^{2} +8 } = \sqrt{\frac{x}{x^{2} +3} }[/tex]

[tex]2^{x^{4} - 10x^{2} +8 } = (\frac{x}{x^{2} +3})^{\frac{1}{2} }[/tex]

[tex]2^{x^{4} - 10x^{2} +8 } = \frac{x^{\frac{1}{2} } }{(x^{2} +3)^{\frac{1}{2} } }[/tex]

[tex]2^{x^{4} - 10x^{2} +8 } = \frac{x^{\frac{1}{2} } }{x^{2.\frac{1}{2} } +3^{\frac{1}{2} } } }[/tex]

[tex]2^{x^{4} - 10x^{2} +8 } = \frac{x^{\frac{1}{2} } }{x + 3^{\frac{1}{2} } } }[/tex]

[tex]2^{x^{4}} : 2^{10x^{2}} . 2^{8} = \frac{x^{\frac{1}{2} } }{x + 3^{\frac{1}{2} } } }[/tex]

[tex]2^{x^{4}} : 2^{10x^{2}} . 2^{8} = \frac{x^{\frac{1}{2} } }{x + 3^{\frac{1}{2} } } }[/tex]

[tex]2^{x^{4}} : 2^{10x^{2}} . (x + 3^{\frac{1}{2} }) = \frac{x^{\frac{1}{2} } }{ 2^{8} } }[/tex]

[tex]\frac{2^{x^{4}} : 2^{10x^{2}} . x + 3^{\frac{1}{2} }}{x^{\frac{1}{2} }} = 2^{8}[/tex]

Answer 2

Verified answer

[tex]\begin{aligned}&{\sf Nilai\ }x{\sf\ yang\ memenuhi\ persamaan}\\ &2^{x^4-10x^2+8}=\sqrt{\dfrac{x}{x^2+3}}\\&{\sf adalah\ \ }\boxed{\,x \in \{\bf1,3\}\,}\ .\end{aligned}[/tex]
 

Penjelasan

Diberikan persamaan:

[tex]2^{x^4-10x^2+8}=\sqrt{\dfrac{x}{x^2+3}}[/tex]

Langkah awal penyelesaian: kuadratkan kedua ruas. Diperoleh:

[tex]4^{x^4-10x^2+8}=\dfrac{x}{x^2+3}[/tex]

Lalu, untuk ruas kiri:

[tex]\begin{aligned}4^{x^4-10x^2+8}&=4^{x^4-x^2-9x^2+9-1}\\&=4^{-1}\cdot4^{x^4-x^2-9x^2+9}\\&=4^{-1}\cdot4^{x^4-x^2-\left(9x^2-9\right)}\\&=4^{-1}\cdot4^{x^2\left(x^2-1\right)-9\left(x^2-1\right)}\\&=\frac{4^{\left(x^2-9\right)\left(x^2-1\right)}}{4}\end{aligned}[/tex]

Maka:

[tex]\begin{aligned}\dfrac{x}{x^2+3}&=\frac{4^{\left(x^2-9\right)\left(x^2-1\right)}}{4}\\4x&=4^{\left(x^2-9\right)\left(x^2-1\right)}(x^2+3)\\0&=4^{\left(x^2-9\right)\left(x^2-1\right)}(x^2+3)-4x\\0&=4^{\left(x^2-9\right)\left(x^2-1\right)}x^2-4x+3\cdot4^{\left(x^2-9\right)\left(x^2-1\right)}\end{aligned}[/tex]

Jumlah akar-akarnya adalah:

[tex]\begin{aligned}x_1+x_2&=\frac{4}{4^{\left(x^2-9\right)\left(x^2-1\right)}}\end{aligned}[/tex]

Hasil kali akar-akarnya adalah:

[tex]\begin{aligned}x_1x_2&=\frac{3\cdot\cancel{4^{\left(x^2-9\right)\left(x^2-1\right)}}}{\cancel{4^{\left(x^2-9\right)\left(x^2-1\right)}}}\\&=3\\&=1\cdot3\ \lor\ (-1)\cdot(-3)\end{aligned}[/tex]

Jelas masih ada pasangan nilai [tex]x[/tex] lain yang memenuhi [tex]x_1x_2=3[/tex].
Namun dapat kita perhatikan bahwa baik untuk [tex]x_1, x_2 \in \{1, 3\}[/tex] maupun untuk [tex]x_1, x_2 \in \{-1, -3\}[/tex]:

[tex]4^{\left(x^2-9\right)\left(x^2-1\right)}=4^0=1[/tex]

yang menyebabkan:

[tex]x_1+x_2=\dfrac{4}{1}=4=1+3[/tex]

Oleh karena itu, nilai [tex]x[/tex] yang memenuhi persamaan di atas adalah:

[tex]\begin{aligned}\boxed{\,x \in \{\bf1,3\}\,}\end{aligned}[/tex]
 
 
[tex]\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}[/tex]