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SNBT Zone : Pangkat (Eksponen)

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Idham Muqoddas
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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.

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
  1. 9
  2. 18
Ganesha Operation
ALTERNATIF PENYELESAIAN
\begin{aligned} 3^x-2\cdot3^{\frac12x+1}+p&=0\\ 3^x-2\cdot3^{\frac12x}\cdot3+p&=0\\ 3^x-6\cdot3^{\frac12x}+p&=0 \end{aligned} Misal {3^{\frac12x}=y} maka
y^2 − 6y + p = 0
\begin{aligned} y_1y_2&=p\\ 3^{\frac12x_1}3^{\frac12x_2}&=p\\ 3^{\frac12x_1+\frac12x_2}&=p\\ 3^{\frac12\left(x_1+x_2\right)}&=p\\ 3^{\frac12\left(2\cdot{^3\negthinspace\log2}+4\right)}&=p\\ 3^{^3\negthinspace\log2+2}&=p\\ 3^{^3\negthinspace\log2}\cdot3^2&=p\\ 2\cdot9&=p\\ p&=\boxed{\boxed{18}} \end{aligned}
Jadi, p = 18.
JAWAB: E

No.

Jika \sqrt[4]{25} +\sqrt[m]{27} = \dfrac2{\sqrt{8-\sqrt{60}}} maka nilai dari m adalah ....
  1. 24
  2. 18
  3. 12
  1. 6
  2. 3
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \sqrt[4]{25} +\sqrt[m]{27}&=\dfrac2{\sqrt{8-\sqrt{60}}}\\ \sqrt[4]{5^2} +\sqrt[m]{27}&=\dfrac2{\sqrt{8-2\sqrt{15}}}\\ \sqrt5 +\sqrt[m]{27}&=\dfrac2{\sqrt{5+3-2\sqrt{5\cdot3}}}\\ \sqrt[m]{27}&=\dfrac2{\sqrt5-\sqrt3}\cdot\dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}-\sqrt5\\ \sqrt[m]{27}&=\dfrac{2\left(\sqrt5+\sqrt3\right)}{5-3}-\sqrt5\\ \sqrt[m]{27}&=\sqrt5+\sqrt3-\sqrt5\\ \sqrt[m]{3^3}&=\sqrt3\\ 3^{\frac3m}&=3^{\frac12}\\ \dfrac3m&=\dfrac12\\ m&=6 \end{aligned}
Jadi, nilai dari m adalah 6.
JAWAB: D

No.

Diberikan a dan b bilangan real a > 1 dan b > 0. Jika ab = ab dan \dfrac{a}b=a^{3b}, maka nilai a adalah
  1. 0
  2. 1
  3. 3
  1. 4
  2. 5
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} ab&=a^b\\ b&=\dfrac{a^b}a\\[3.7pt] &=a^{b-1} \end{aligned} \begin{aligned} \dfrac{a}b&=a^{3b}\\[3.7pt] \dfrac{a}{a^{b-1}}&=a^{3b}\\[3.7pt] a^{1-(b-1)}&=a^{3b}\\ a^{1-b+1}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ 2&=4b\\ b&=\dfrac12 \end{aligned} \begin{aligned} ab&=a^b\\ a\left(\dfrac12\right)&=a^{\frac12}\\[3.7pt] \dfrac12a&=\sqrt{a}\\[3.7pt] \dfrac14a^2&=a\\[3.7pt] a^2&=4a\\ a&=4 \end{aligned}
Jadi, nilai a adalah 4.
JAWAB:

No.

Jika A2x = 3, maka nilai \dfrac{A^{5x}-A^{-5x}}{A^{3x}+A^{-3x}}=
  1. \dfrac{121}{42}
  2. \dfrac{120}{41}
  3. \dfrac{42}{121}
  1. \dfrac{41}{120}
  2. \dfrac{14}{21}
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \dfrac{A^{5x}-A^{-5x}}{A^{3x}+A^{-3x}}{\color{red}{\cdot\dfrac{A^{5x}}{A^{5x}}}}&=\dfrac{A^{10x}-A^0}{A^{8x}+A^{2x}}\\[3.7pt] &=\dfrac{\left(A^{2x}\right)^5-1}{\left(A^{2x}\right)^4+A^{2x}}\\[3.7pt] &=\dfrac{3^5-1}{3^4+3}\\[3.7pt] &=\dfrac{243-1}{81+3}\\[3.7pt] &=\dfrac{242}{84}\\ &=\boxed{\boxed{\dfrac{121}{42}}} \end{aligned}
Jadi, nilai \dfrac{A^{5x}-A^{-5x}}{A^{3x}+A^{-3x}}=\dfrac{121}{42}.
JAWAB: A

No.

Misalkan x, y, r, s > 1 adalah bilangan real yang memenuhi persamaan x7y5 = r dan x4y3 = s. Jika x = rasb dan y = rcsd, maka nilai dari a + b + c + d adalah ....
  1. 0
  2. 1
  3. 2
  1. 3
  2. 4
TO SIMAK UI
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \dfrac{\left(x^7y^5\right)^3}{\left(x^4y^3\right)^5}&=\dfrac{r^3}{s^5}\\[3.7pt] \dfrac{x^{21}y^{15}}{x^{20}y^{15}}&=r^3s^{-5}\\[3.7pt] x&=r^3s^{-5} \end{aligned}\)
a = 3, b = −5

\(\begin{aligned} \dfrac{\left(x^7y^5\right)^4}{\left(x^4y^3\right)^7}&=\dfrac{r^4}{s^7}\\[3.7pt] \dfrac{x^{28}y^{20}}{x^{28}y^{21}}&=\dfrac{r^4}{s^7}\\[3.7pt] \dfrac1y&=\dfrac{r^4}{s^7}\\[3.7pt] y&=\dfrac{s^7}{r^4}\\[3.7pt] &=r^{-4}s^7 \end{aligned}\)
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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