Latihan Soal SBMPTN Bab: Eksponen Dan Bentuk Akar Lengkap Dengan Pembahasan

Soal UTBK Eksponen Dan Bentuk Akar SMA/SMK

01

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Jika $\sqrt[3]{4^{x+1}}=2\sqrt{8^{x}}$, maka $x=...$

A. $-\frac{2}{5}$

B. $-\frac{1}{5}$

C. $\frac{1}{5}$

D. $\frac{2}{5}$

D. $1$

Pembahasan:

Show

\begin{aligned} \sqrt[3]{4^{x+1}} &= 2\sqrt{8^{x}} \\ \sqrt[3]{2^{x+1}} &= 2\sqrt{2^{3x}} \\ 2^{\frac{2x+2}{3}}&= 2(2^{\frac{3}{2}x}) \\ 2^{\frac{2x+2}{3}}&= 2^{1+\frac{3}{2}x} \\ \frac{2x+2}{3} &= 1+\frac{3}{2}x \\ 4x+4 &= 6+9x \\ 5x &= -1\\ x&=-\frac{1}{5} \end{aligned}

JAWAB : B

02

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Jika $4^{x}-4^{x-1}=6$, maka $(2x)^{x}$ sama dengan

A. $3$

B. $3\sqrt{3}$

C. $9$

D. $9\sqrt{3}$

D. $27$

Pembahasan:

Show

\begin{aligned} 4^{x}-4^{x-1} &= 6 &\rightarrow 4^{x}-\frac{4^{x}}{4}&=6 \\ 3\times 4^{x} &= 24 &\rightarrow 4^{x}&=8 \\ 2^{2x}&= 2^{3} &\rightarrow x&=\frac{3}{2} \\ \end{aligned} jadi, $(2x)^{x} = (3)^{\frac{3}{2}}=3\times 3^{\frac{1}{2}}=3\sqrt{3}$

JAWAB : B

03

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Nilai $x$ yang memenuhi persamaan $\sqrt[3]{4^{5-x}}=\frac{1}{2^{2x-1}}$ adalah...

A. $-4$

B. $-1$

C. $-\frac{1}{2}$

D. $\frac{1}{4}$

D. $2$

Pembahasan:

Show

\begin{aligned} \frac{\sqrt[3]{4^{5-x}}}{8} &= \frac{1}{2x+1}\Leftrightarrow \frac{2^{\frac{10-2x}{3}}}{2^{3}} = 2^{-2x-1} \\ 2^{\frac{10-2x}{3}-3} &= 2^{-2x-1} \\ \frac{10-2x}{3} - 3&= -2x-1 \\ 10-2x-9 &=-6-3 \\ 4x &= -4\Leftrightarrow x=-1 \end{aligned}

JAWAB : B

04

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Jika $\frac{\frac{1}{2}-\frac{1}{\sqrt{5}}}{\frac{1}{2}+\frac{1}{\sqrt{5}}} = a+b\sqrt{5}$

Maka $a+b$

A. $-4$

B. $5$

C. $-\frac{1}{2}$

D. $\frac{1}{4}$

D. $2$

Pembahasan:

Show

$a+b\sqrt{5}=\frac{\frac{1}{2}-\frac{1}{5}}{\frac{1}{2}+\frac{1}{\sqrt{5}}}=\frac{\frac{\sqrt{5}-2}{2\sqrt{5}}}{\frac{\sqrt{5}+2}{2\sqrt{5}}}=\frac{\sqrt{5}-2}{5-4}$

Rasionalkan penyebutnya

\begin{flalign} a+b\sqrt{5}&= \frac{\sqrt{5}-2}{\sqrt{5}-2}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{9-4\sqrt{5}}{5-4} &\\ a+b\sqrt{5}&= 9-4\sqrt{5} &\\ \end{flalign} \begin{flalign} \text {Jadi nilai a = 9 dan b = -4,} &\\ \text {Sehingga a+b = 9+(-4) = 5} &\\ \end{flalign} JAWAB : B

05

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Bentuk akar $\frac{(a^{\frac{5}{3}}b^{\frac{1}{2}}-a^{\frac{2}{2}}b^{\frac{3}{2}})}{(a^{\frac{7}{2}}b^{\frac{1}{2}}-a^{\frac{3}{2}}b)}$

Maka $a+b$

A. $\sqrt{a}-\sqrt{b}$

B. $\sqrt{a}+\sqrt{b}$

C. $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$

D. $\frac{\sqrt{a}-\sqrt{b}}{a-b}$

E. $\frac{\sqrt{a}+\sqrt{b}}{a+b}$

Pembahasan:

Show

\begin{aligned} \frac{(a^{\frac{5}{3}}b^{\frac{1}{2}}-a^{\frac{2}{2}}b^{\frac{3}{2}})}{(a^{\frac{7}{2}}b^{\frac{1}{2}}-a^{\frac{3}{2}}b)}= \frac{b^{\frac{1}{2}}a^{\frac{2}{3}}\\(a-b)}{ a^{\frac{2}{3}}b^{\frac{1}{2}}\\(\sqrt{a}+\sqrt{b})} \\ = \frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{ (\sqrt{a}-\sqrt{b})} = \sqrt{a}+\sqrt{b} \end{aligned}

JAWAB : B

06

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Jika $r = \frac{20\sqrt{2} - 25}{(10+20\sqrt{2})(2-\sqrt{2})}$, maka $(4r-2)^2 = ....$

A. $5$

B. $4$

C. $3$

D. $2$

E. $1$

Pembahasan:

Show

\begin{flalign} r & = \frac{20\sqrt{2} - 25}{(10+20\sqrt{2})(2-\sqrt{2})} \\ & = \frac{5(4\sqrt{2} - 5)}{10(1+2\sqrt{2})(2-\sqrt{2})} \, \, \, \, \, \, \, \text{(bagi 5)} &\\ & = \frac{(4\sqrt{2} - 5)}{2(1+2\sqrt{2})(2-\sqrt{2})} \, \, \, \, \, \, \, \text{(kalikan penyebutnya)} &\\ & = \frac{(4\sqrt{2} - 5)}{2(3\sqrt{2} - 2)} \, \, \, \, \, \, \, \text{(kalikan sekawan)} &\\ & = \frac{(4\sqrt{2} - 5)}{2(3\sqrt{2} - 2)} . \frac{3\sqrt{2} + 2}{3\sqrt{2} + 2} &\\ & = \frac{14 - 7\sqrt{2}}{2(9.2- 4 )} = \frac{7(2 - \sqrt{2})}{2.14} = \frac{(2 - \sqrt{2})}{4} \end{flalign}

Menentukan nilai $(4r-2)^2$

\begin{flalign} (4r-2)^2 & = \left( 4.\frac{(2 - \sqrt{2})}{4} - 2 \right)^2 &\\ & = \left( (2 - \sqrt{2}) - 2 \right)^2 &\\ & = \left( - \sqrt{2} \right)^2 &\\ & = 2 \end{flalign} JAWAB : D

07

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Bentuk sederhana dari $\dfrac{\left(\sqrt{3}+\sqrt{7} \right)\left( \sqrt{3}-\sqrt{7} \right)}{2\sqrt{5}-4\sqrt{2}}$ adalah...

A. $\frac{2}{3} \left(\sqrt{5}+2\sqrt{2} \right)$

B. $\frac{2}{3} \left(2\sqrt{2}-\sqrt{5} \right)$

C. $-\frac{2}{3} \left( 2\sqrt{5}+ 4\sqrt{2} \right)$

D. $-\frac{4}{9} \left( 2\sqrt{5}+ 4\sqrt{2} \right)$

E. $-\frac{4}{9} \left( 2\sqrt{5}-\sqrt{2} \right)$

Pembahasan:

Show

\begin{flalign} & \dfrac{\left(\sqrt{3}+\sqrt{7} \right)\left( \sqrt{3}-\sqrt{7} \right)}{2\sqrt{5}-4\sqrt{2}} &\\ & = \dfrac{3-7}{2 \left( \sqrt{5}-2\sqrt{2} \right) } \times \dfrac{\sqrt{5}+2\sqrt{2}}{\sqrt{5}+2\sqrt{2}} &\\ & = \dfrac{-4 \left( \sqrt{5} + 2\sqrt{2} \right)}{2 \left( 5- 8 \right) } &\\ & = \dfrac{-4 \left( \sqrt{5} + 2\sqrt{2} \right)}{-6} \\ & = \dfrac{2}{3} \left( \sqrt{5} + 2\sqrt{2} \right) &\\ \end{flalign} JAWAB : A

08

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$\dfrac{5 \left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)^{3}}{\left( 2\sqrt{2}-\sqrt{3} \right)}=\cdots$

A. $\sqrt{3}-\sqrt{2}$

B. $3\sqrt{3}-2\sqrt{2}$

C. $2\sqrt{2}-3\sqrt{3}$

D. $3\sqrt{2}-2\sqrt{3}$

E. $4\sqrt{2}-3\sqrt{3} $

Pembahasan:

Show

\begin{flalign} & \dfrac{5 \left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)^{3}}{\left( 2\sqrt{2}-\sqrt{3} \right)} &\\ & = \dfrac{5 \left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)^{2}}{\left( 2\sqrt{2}-\sqrt{3} \right)} &\\ & = \dfrac{5 \left( 3-2 \right)\left( \sqrt{3}-\sqrt{2} \right)^{2}}{\left( 2\sqrt{2}-\sqrt{3} \right)} &\\ & = \dfrac{5 \left( 3+2-2\sqrt{6} \right)}{\left( 2\sqrt{2}-\sqrt{3} \right)} &\\ & = \dfrac{25-10\sqrt{6}}{\left( 2\sqrt{2}-\sqrt{3} \right)} \times \dfrac{\left( 2\sqrt{2}+\sqrt{3} \right)}{\left( 2\sqrt{2}+\sqrt{3} \right)} &\\ & = \dfrac{50\sqrt{2}+25\sqrt{3}-20\sqrt{12}-10\sqrt{18}}{\left( 8-3 \right)} &\\ & = \dfrac{50\sqrt{2}+25\sqrt{3}-40\sqrt{3}-30\sqrt{2}}{5} &\\ & = \dfrac{20\sqrt{2}-15\sqrt{3}}{5} &\\ & = 4\sqrt{2}-3\sqrt{3} \end{flalign} JAWAB : E

09

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Nilai x yang memenuhi persamaan $3^{2x+3}=\sqrt[3]{27^{x+5}}$

A. $-2$

B. $-1$

C. $0$

D. $1$

E. $2$

Pembahasan:

Show

\begin{flalign} 3^{2x+3} &=\sqrt[3]{27^{x+5}} &\\ 3^{2x+3} &=27^{\dfrac{x+5}{3}} &\\ 3^{2x+3} &=(3^{3})^{\dfrac{x+5}{3}} &\\ 3^{2x+3} &=3^{x+5} &\\ & \Rightarrow 2x+3=x+5 &\\ & \Rightarrow 2x-x=5-3 &\\ & \Rightarrow x=2 \end{flalign} JAWAB : E

10

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Nilai $1-x$ yang memenuhi persamaan $\sqrt{8^{3-x}}=4 \cdot 2^{1-2x}$

A. $-4$

B. $-3$

C. $-2$

D. $3$

E. $4$

Pembahasan:

Show

\begin{flalign} \sqrt{8^{3-x}} &= 4 \cdot 2^{1-2x} &\\ 8^{\dfrac{3-x}{2}} &= 2^{2} \cdot 2^{1-2x} &\\ 2^{ \dfrac{3(3-x)}{2}} &= 3-2x &\\ 9-3x &= 6-4x &\\ 4x-3x &= 6-9 &\\ x &= -3 &\\ 1- x &= 1-(-3) =4 \end{flalign} JAWAB : E