SBMPTN Zone : Pangkat (Eksponen)

Berikut ini adalah kumpulan soal mengenai Pangkat (Eksponen) tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Jika x_1, x_2 adalah akar-akar 3^x-2\cdot3^{\frac12x+1}+p=0 dan x_1+x_2=2\cdot{^3\negthinspace\log2}+4 maka p= ....

2. 
   
3. 
   

5. 
   

Ganesha Operation

$$\begin{array}{rl}{3}^x-2\cdot 3^{\frac{1}{2}x+1}+p& =0\\ 3^x-2\cdot 3^{\frac{1}{2}x}\cdot 3+p& =0\\ 3^x-6\cdot 3^{\frac{1}{2}x}+p& =0\end{array}$$
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 ....

2. 
   
3. 
   

5. 
   

sumber

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

Jadi,

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No. 3

Diberikan a dan b bilangan real a\gt1 dan b\gt0. Jika ab=a^b dan \dfrac{a}b=a^{3b}, maka nilai a adalah

1. 0
2. 1
   
3. 3
   

4. 4
5. 5
   

$$\begin{array}{rl}ab& =a^b\\ b& ={\displaystyle \frac{a^b}{a}}\\ & =a^{b-1}\end{array}$$

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

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

Jadi, a=4.
JAWAB: D

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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}

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

Jadi,