sma kelas 10 tentukan nilai dari x.... Question from @anonimczh

xcvi

Menentukan nilai x dari:

$\large \begin{aligned}\sf 1. \: \: \: \: \: \: \: \: \: \sqrt{ {8}^{2x -6 } } &=\sf \frac{1}{16} \\ \sf \sqrt{ { {(2}^{3} )}^{2x - 6} } &=\sf \frac{1}{ {2}^{4} } \\ \sf \sqrt{ {2}^{6x - 18} } &= \sf {2}^{ - 4} \\ \sf {2}^{3x - 9} &=\sf {2}^{ - 4} \\ \sf 3x - 9&=\sf - 4\\ \sf 3x&=\sf - 4 + 9\\ \sf 3x&=\sf 5\\ \sf x&=\sf \frac{5}{3} \end{aligned}$

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$\large \begin{aligned}\sf 2. \: \: \: \: \: \: \: \: \: \: \: \: \frac{ {9}^{x - 1} }{ {27}^{x - 1} } &=\sf 3\\ \sf \frac{ { {(3}^{2} )}^{x - 1} }{ { {(3}^{3} )}^{x - 1} } &=\sf 3\\ \sf \frac{ {3}^{2x - 2} }{ {3}^{3x - 3} } &= \sf 3\\ \sf {3}^{2x - 3x - 2 - ( - 3)} &=\sf 3\\ \sf {3}^{ - x + 1} &=\sf3\\ \sf - x + 1&=\sf 1\\ \sf x&=\sf1 - 1\\ \sf x&=\sf0\end{aligned}$

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$\large \begin{aligned}\sf 3. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {5}^{ {x}^{2} - 2x - 18 } &=\sf \frac{1}{125} \\ \sf {5}^{ {x}^{2} - 2x - 18 }&=\sf \frac{1}{ {5}^{3} } \\ \sf {5}^{ {x}^{2} - 2x - 18 }&= \sf {5}^{ - 3} \\ \sf {x}^{2} - 2x - 18 &=\sf - 3\\ \sf {x}^{2} - 2x - 18 + 3 &=\sf0\\ \sf {x}^{2} - 2x - 15 &=\sf0\\ \sf (x + 3)(x - 5) &=\sf0 \\ \sf sehingga: \\\sf x = - 3 \: atau \: x &=\sf 5\end{aligned}$

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» Detail Jawaban

⚜ Mapel: Matematika

⚜ Kelas: 10

⚜ Bab: Eksponen

⚜ Kode Kategorisasi: 10.2.1.1

#SamaSamaBelajar

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Jawab:
$\displaystyle(1.)~~\bf x = \frac{5}{3}$
(2.) x = 0
(3.) x = {-3, 5}

Penjelasan dengan langkah-langkah:
Nomor 1
$\displaystyle\sf\sqrt{8^{2x-6}}=\frac{1}{16}~~~8^{\frac{1}{2}(2x-6)}=\frac{1}{2^4}\\\\(2^3)^{\frac{1}{2}(2x-6)}=2^{-4}\\\\(2^3)^{\frac{2x}{2}-\frac{6}{2}}=2^{-4}~~(2^3)^{x-3}=2^{-4}$
3(x-3) = -4 3x-9 = -4 3x=9-4
3x = 5
$\displaystyle\therefore~~\bf x = \frac{5}{3}$

Nomor 2
$\displaystyle\sf\frac{9^{x-1}}{27^{x-1}}=3~~~~\frac{(3^2)^{x-1}}{(3^3)^{x-1}}=3\\\\\frac{3^{2x-2}}{3^{3x-3}}=3^1~~~~3^{2x-2-(3x-3)}=3^1$
2x-2-(3x-3) = 1
2x-2-3x+3 = 1
-x+1 = 1
x = 1-1
x = 0

Nomor 3
$\displaystyle\rm5^{x^2-2x-18}=\frac{1}{125}~~~5^{x^2-2x-18}=\frac{1}{5^3}\\\\5^{x^2-2x-18}=5^{-3}$
x²-2x-18 = -3 x²-2x-18+3 = 0
x²-2x-15 = 0 Ax²+Bx+C = 0
A = 1, B = -2, C = -15
p dan q akar-akar penyelesaian
$\rm p+q = ~^{-B}/_A ~~~ pq =~ ^{C}/_A$
p+q = ⁻⁽⁻²⁾/₁ pq = ⁻¹⁵/₁
p+q = 2 pq = -15
p = 5, q = -3
Maka x = {-3, 5}

(xcvi)

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