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{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

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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{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,

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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{aligned}
ab&=a^b\\
b&=\dfrac{a^b}a\\[8pt]
&=a^{b-1}
\end{aligned}

\begin{aligned}
\dfrac{a}b&=a^{3b}\\[8pt]
\dfrac{a}{a^{b-1}}&=a^{3b}\\[8pt]
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}\\[8pt]
\dfrac12a&=\sqrt{a}\\[8pt]
\dfrac14a^2&=a\\[8pt]
a^2&=4a\\
a&=4
\end{aligned}

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{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}}\\[8pt]
&=\dfrac{\left(A^{2x}\right)^5-1}{\left(A^{2x}\right)^4+A^{2x}}\\[8pt]
&=\dfrac{3^5-1}{3^4+3}\\[8pt]
&=\dfrac{243-1}{81+3}\\[8pt]
&=\dfrac{242}{84}\\
&=\boxed{\boxed{\dfrac{121}{42}}}
\end{aligned}

Jadi,