SNBT Zone : Pangkat (Eksponen)

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Berikut ini adalah kumpulan soal mengenai Pangkat (Eksponen). Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSNBTHOTS

No. 1

Jika x₁, x₂ adalah akar-akar {3^x-2\cdot3^{\frac12x+1}+p=0} dan x₁ + x₂ = 2⋅³log 2 + 4 maka p = ....

1. 2
2. 3
3. 8

4. 9
5. 18

Ganesha Operation

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{3}^x-2\cdot 3^{\frac{1}{2}x+1}+p& =0\\ 3^x-2\cdot 3^{\frac{1}{2}x}\cdot 3+p& =0\\ 3^x-6\cdot 3^{\frac{1}{2}x}+p& =0\end{array}$$ Misal {3^{\frac12x}=y} maka
y^2 − 6y + p = 0

$$\begin{array}{rl}{y}_1{y}_2& =p\\ 3^{\frac{1}{2}{x}_1}{3}^{\frac{1}{2}{x}_2}& =p\\ 3^{\frac{1}{2}{x}_1+\frac{1}{2}{x}_2}& =p\\ 3^{\frac{1}{2}(x_1+x_2)}& =p\\ 3^{\frac{1}{2}(2\cdot {}^3\log 2+4)}& =p\\ 3^{{}^3\log 2+2}& =p\\ 3^{{}^3\log 2}\cdot 3^2& =p\\ 2\cdot 9& =p\\ p& =\boxed{{\displaystyle \boxed{{\displaystyle 18}}}}\end{array}$$

Jadi, p = 18.
JAWAB: E

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No. 2

Jika \sqrt[4]{25} +\sqrt[m]{27} = \dfrac2{\sqrt{8-\sqrt{60}}} maka nilai dari m adalah ....

1. 24
2. 18
3. 12

4. 6
5. 3

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\sqrt[4]{25}+\sqrt[m]{27}& ={\displaystyle \frac{2}{\sqrt{8-\sqrt{60}}}}\\ \sqrt[4]{5^2}+\sqrt[m]{27}& ={\displaystyle \frac{2}{\sqrt{8-2\sqrt{15}}}}\\ \sqrt{5}+\sqrt[m]{27}& ={\displaystyle \frac{2}{\sqrt{5+3-2\sqrt{5\cdot 3}}}}\\ \sqrt[m]{27}& ={\displaystyle \frac{2}{\sqrt{5}-\sqrt{3}}}\cdot {\displaystyle \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}}-\sqrt{5}\\ \sqrt[m]{27}& ={\displaystyle \frac{2(\sqrt{5}+\sqrt{3})}{5-3}}-\sqrt{5}\\ \sqrt[m]{27}& =\sqrt{5}+\sqrt{3}-\sqrt{5}\\ \sqrt[m]{3^3}& =\sqrt{3}\\ 3^{\frac{3}{m}}& =3^{\frac{1}{2}}\\ {\displaystyle \frac{3}{m}}& ={\displaystyle \frac{1}{2}}\\ m& =6\end{array}$$

Jadi, nilai dari m adalah 6.
JAWAB: D

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No. 3

Diberikan a dan b bilangan real a > 1 dan b > 0. Jika ab = a^b dan \dfrac{a}b=a^{3b}, maka nilai a adalah

1. 0
2. 1
3. 3

4. 4
5. 5

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}ab& =a^b\\ b& ={\displaystyle \frac{a^b}{a}}\\ & =a^{b-1}\end{array}$$ $$\begin{array}{rl}{\displaystyle \frac{a}{b}}& =a^{3b}\\ {\displaystyle \frac{a}{a^{b-1}}}& =a^{3b}\\ a^{1-(b-1)}& =a^{3b}\\ a^{1-b+1}& =a^{3b}\\ a^{2-b}& =a^{3b}\\ 2-b& =3b\\ 2& =4b\\ b& ={\displaystyle \frac{1}{2}}\end{array}$$ $$\begin{array}{rl}ab& =a^b\\ a\left({\displaystyle \frac{1}{2}}\right)& =a^{\frac{1}{2}}\\ {\displaystyle \frac{1}{2}}a& =\sqrt{a}\\ {\displaystyle \frac{1}{4}}{a}^2& =a\\ a^2& =4a\\ a& =4\end{array}$$

Jadi, nilai a adalah 4.
JAWAB:

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No. 4

Jika A^2x = 3, maka nilai \dfrac{A^{5x}-A^{-5x}}{A^{3x}+A^{-3x}}=

1. \dfrac{121}{42}
2. \dfrac{120}{41}
3. \dfrac{42}{121}

4. \dfrac{41}{120}
5. \dfrac{14}{21}

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{A^{5x}-A^{-5x}}{A^{3x}+A^{-3x}}}\text{color}red\cdot {\displaystyle \frac{A^{5x}}{A^{5x}}}& ={\displaystyle \frac{A^{10x}-A^0}{A^{8x}+A^{2x}}}\\ & ={\displaystyle \frac{{\left(A^{2x}\right)}^5-1}{{\left(A^{2x}\right)}^4+A^{2x}}}\\ & ={\displaystyle \frac{3^5-1}{3^4+3}}\\ & ={\displaystyle \frac{243-1}{81+3}}\\ & ={\displaystyle \frac{242}{84}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{121}{42}}}}}}\end{array}$$

Jadi, nilai \dfrac{A^{5x}-A^{-5x}}{A^{3x}+A^{-3x}}=\dfrac{121}{42}.
JAWAB: A

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No. 5

Misalkan x, y, r, s > 1 adalah bilangan real yang memenuhi persamaan x⁷y⁵ = r dan x⁴y³ = s. Jika x = r^as^b dan y = r^cs^d, maka nilai dari a + b + c + d adalah ....

1. 0
2. 1
3. 2

4. 3
5. 4

TO SIMAK UI

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{{\left(x^7{y}^5\right)}^3}{{\left(x^4{y}^3\right)}^5}}& ={\displaystyle \frac{r^3}{s^5}}\\ {\displaystyle \frac{x^{21}{y}^{15}}{x^{20}{y}^{15}}}& =r^3{s}^{-5}\\ x& =r^3{s}^{-5}\end{array}$
a = 3, b = −5

$\begin{array}{rl}{\displaystyle \frac{{\left(x^7{y}^5\right)}^4}{{\left(x^4{y}^3\right)}^7}}& ={\displaystyle \frac{r^4}{s^7}}\\ {\displaystyle \frac{x^{28}{y}^{20}}{x^{28}{y}^{21}}}& ={\displaystyle \frac{r^4}{s^7}}\\ {\displaystyle \frac{1}{y}}& ={\displaystyle \frac{r^4}{s^7}}\\ y& ={\displaystyle \frac{s^7}{r^4}}\\ & =r^{-4}{s}^7\end{array}$

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c = −4, d = 7

a + b + c + d = 3 + (−5) + (−4) + 7 = 1

Jadi, a + b + c + d = 1.
JAWAB: B

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