Soal Bilangan Berpangkat Pecahan dan Pembahasan

Hallo...! Pengunjung setia Catatan Matematika, kali ini Bang RP (Reikson Panjaitan, S.Pd) berbagi Kumpulan Soal Bilangan Berpangkat Pecahan berserta pembahasannya. Ayo... manfaatkan website Catatan Matematika ini untuk belajar matematika secara online.

Tata Cara Belajar:
Cobalah mengerjakan soal-soal yang tersedia secara mandiri. Setelah itu cek jawaban kamu dengan pembahasan yang telah disediakan, dengan cara:
klik "LIHAT/TUTUP:".

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Soal No. 1

Bentuk $\sqrt{5}$ sama nilainya dengan ….
A. $2^{\frac{1}{5}}$
B. $5^{\frac{1}{2}}$
C. ${5^2}$
D. ${2^{5}}$
E. $\frac{5}{2}$

Penyelesaian: Lihat/Tutup

Ingat: $\sqrt[n]{a^m}=a^{\frac{m}{n}}$ maka $\sqrt{5}=5^{\frac{1}{2}}$
Jawaban: B

Soal No. 2

Nilai dari $16^{\frac{1}{2}}$ sama dengan ….
A. 8
B. 6
C. 4
D. 2
E. 1

Penyelesaian: Lihat/Tutup

$\begin{align}16^{\frac{1}{2}} &= (4^2)^{\frac{1}{2}} \\ &= 4^{2.\frac{1}{2}} \\ &= 4 \end{align}$
Jawaban: C

Soal No. 3

$81^{-\frac{1}{2}}$ = ….
A. 9
B. 3
C. 1
D. $\frac{1}{3}$
E. $\frac{1}{9}$

Penyelesaian: Lihat/Tutup

$\begin{align}81^{-\frac{1}{2}} &= (9^2)^{-\frac{1}{2}} \\ &= 9^{2.\left( -\frac{1}{2} \right)} \\ &= 9^{-1} \\ &= \frac{1}{9} \end{align}$
Jawaban: E

Soal No. 4

Nilai dari $\sqrt{2^4}$ sama dengan ….
A. 2
B. 4
C. 6
D. 8
E. 16

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt{2^4} &= 2^{\frac{4}{2}} \\ &= {2^2} \\ &= 4 \end{align}$
Jawaban: B

Soal No. 5

$(\sqrt{81})^{\frac{1}{2}}$ = ….
A. 3
B. 6
C. 9
D. 12
E. 36

Penyelesaian: Lihat/Tutup

$\begin{align}(\sqrt{81})^{\frac{1}{2}} &= (\sqrt{3^4})^{\frac{1}{2}} \\ &= \left( 3^{\frac{4}{2}} \right)^{\frac{1}{2}} \\ &= 3^{\frac{4}{2}.\frac{1}{2}} \\ &= 3 \end{align}$
Jawaban: A

Soal No. 6

Hasil dari $(64)^{-\frac{1}{3}}$ adalah ….
A. $\frac{1}{8}$
B. $\frac{1}{4}$
C. $\frac{1}{2}$
D. 8
E. 4

Penyelesaian: Lihat/Tutup

$\begin{align}(64)^{-\frac{1}{3}} &= (4^3)^{-\frac{1}{3}} \\ &= 4^{3.\left( -\frac{1}{3} \right)} \\ &= 4^{-1} \\ &= \frac{1}{4} \end{align}$
Jawaban: B

Soal No. 7

$\left( \sqrt{\sqrt{4}} \right)^2$ = ….
A. 1
B. 2
C. 3
D. 4
E. 8

Penyelesaian: Lihat/Tutup

$\left( \sqrt{\sqrt{4}} \right)^2=\left( \sqrt{2} \right)^2=2$
Jawaban: B

Soal No. 8

Nilai dari $27^{\frac{4}{3}}+27^{-\frac{4}{3}}+27^0$ adalah ….
A. 80
B. $80\frac{1}{81}$
C. $81\frac{1}{80}$
D. $81\frac{1}{82}$
E. $82\frac{1}{81}$

Penyelesaian: Lihat/Tutup

$\begin{align}27^{\frac{4}{3}}+27^{-\frac{4}{3}}+27^0 &= (3^3)^{\frac{4}{3}}+(3^3)^{-\frac{4}{3}}+27^0 \\ &= 3^{3.\frac{4}{3}}+3^{3.\left( -\frac{4}{3} \right)}+1 \\ &= 3^4+3^{-4}+1 \\ &= 3^4+\frac{1}{3^4}+1 \\ &= 81+\frac{1}{81}+1 \\ &= 82\frac{1}{81} \end{align}$
Jawaban: E

Soal No. 9

$\sqrt{2}$, $\sqrt[3]{4}$, $\sqrt[4]{6}$ ditulis dalam urutan naik adalah ….
A. $\sqrt{2}$, $\sqrt[3]{4}$, $\sqrt[4]{6}$
B. $\sqrt{2}$, $\sqrt[4]{6}$, $\sqrt[3]{4}$
C. $\sqrt[3]{4}$, $\sqrt[4]{6}$, $\sqrt{2}$
D. $\sqrt[4]{6}$, $\sqrt[3]{4}$, $\sqrt{2}$
E. $\sqrt[4]{6}$, $\sqrt{2}$, $\sqrt[3]{4}$

Penyelesaian: Lihat/Tutup

$\sqrt{2}=2^{\frac{1}{2}}$, $\sqrt[3]{4}=4^{\frac{1}{3}}$, dan $\sqrt[4]{6}={6^{\frac{1}{4}}}$
Masing-masing pangkatnya pecahan, yang penyebutnya 2, 3 dan 4.
KPK dari (2, 3 dan 4) adalah 12.
Masing-masing kita pangkatnya dengan 12 diperoleh:
$\sqrt{2}=2^{\frac{1}{2}}\to {{\left( {2^{\frac{1}{2}}} \right)}^{12}}={2^{\frac{1}{2}.12}}=2^6=32$
$\sqrt[3]{4}=4^{\frac{1}{3}}\to \left( 4^{\frac{1}{3}} \right)^{12}=4^4=256$
$\sqrt[4]{6}=6^{\frac{1}{4}}=\left( 6^{\frac{1}{4}} \right)^{12}=6^3=216$
32 < 216 < 256 maka urutan naik bilangan tersebut adalah $\sqrt{2}$, $\sqrt[4]{6}$, $\sqrt[3]{4}$
Jawaban: B

Soal No. 10

$\sqrt[5]{ab^2\sqrt[3]{{b^2}\sqrt{ab}}}$ = ….
A. $\sqrt[15]{{a^7}{b^9}}$
B. $\sqrt[15]{{a^7}{b^{11}}}$
C. $\sqrt[1305]{{a^7}{b^7}}$
D. $\sqrt[30]{{a^{17}}{b^7}}$
E. $\sqrt[30]{{a^7}{b^{17}}}$

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt[5]{ab^2\sqrt[3]{{b^2}\sqrt{ab}}} &= \sqrt[5]{ab^2\sqrt[3]{{b^2}{{(ab)}^{\frac{1}{2}}}}} \\ &= \sqrt[5]{ab^2{{\left( {b^2}{{(ab)}^{\frac{1}{2}}} \right)}^{\frac{1}{3}}}} \\ &= {{\left( ab^2{{\left( {b^2}{{(ab)}^{\frac{1}{2}}} \right)}^{\frac{1}{3}}} \right)}^{\frac{1}{5}}} \\ &= {a^{\frac{1}{5}}}.{b^{\frac{2}{5}}}{{\left( {b^2}{{(ab)}^{\frac{1}{2}}} \right)}^{\frac{1}{15}}} \\ &= {a^{\frac{1}{5}}}.{b^{\frac{2}{5}}}.{b^{\frac{2}{15}}}.{a^{\frac{1}{30}}}{b^{\frac{1}{30}}} \\ &= {a^{\frac{1}{5}+\frac{1}{30}}}.{b^{\frac{2}{5}+\frac{2}{15}+\frac{1}{30}}} \\ &= {a^{\frac{6}{30}+\frac{1}{30}}}.{b^{\frac{12}{30}+\frac{4}{30}+\frac{1}{30}}} \\ &= {a^{\frac{7}{30}}}.{b^{\frac{17}{30}}} \\ &= {{\left( {a^7}.{b^{17}} \right)}^{\frac{1}{30}}} \\ &= \sqrt[30]{{a^7}{b^{17}}} \end{align}$
Jawaban: E

Soal No. 11

Jika $a=32$ dan $b=27$, maka nilai dari ${a^{\frac{1}{5}}}+{b^{\frac{1}{3}}}$ adalah ….
A. $\frac{1}{5}$
B. $\frac{1}{6}$
C. 5
D. 6
E. 8

Penyelesaian: Lihat/Tutup

$\begin{align}{a^{\frac{1}{5}}}+{b^{\frac{1}{3}}} &= {32^{\frac{1}{5}}}+{27^{\frac{1}{3}}} \\ &= {{({2^5})}^{\frac{1}{5}}}+{{({3^3})}^{\frac{1}{3}}} \\ &= {2^{5.\frac{1}{5}}}+{3^{3.\frac{1}{3}}} \\ &= 2+3 \\ &= 5 \end{align}$
Jawaban: C

Soal No. 12

Nilai dari $\frac{{36^{\frac{1}{2}}}}{{27^{\frac{2}{3}}}-{{\left( \frac{1}{2} \right)}^{-2}}}$ adalah ….
A. $\frac{6}{13}$
B. $\frac{13}{6}$
C. $\frac{24}{37}$
D. $\frac{24}{35}$
E. $\frac{6}{5}$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{{36^{\frac{1}{2}}}}{{27^{\frac{2}{3}}}-{{\left( \frac{1}{2} \right)}^{-2}}} &= \frac{{{({6^2})}^{\frac{1}{2}}}}{{{({3^3})}^{\frac{2}{3}}}-{{({2^{-1}})}^{-2}}} \\ &= \frac{{6^{2.\frac{1}{2}}}}{{3^{3.\frac{2}{3}}}-{2^{-1.(-2)}}} \\ &= \frac{6}{{3^2}-{2^2}} \\ &= \frac{6}{9-4} \\ &= \frac{6}{5} \end{align}$
Jawaban: E

Soal No. 13

Nilai dari ${(243)^{\frac{2}{5}}}{(64)^{-\frac{1}{2}}}$ = ….
A. $-\frac{27}{8}$
B. $-\frac{9}{8}$
C. $\frac{9}{8}$
D. $\frac{18}{8}$
E. $\frac{27}{8}$

Penyelesaian: Lihat/Tutup

$\begin{align}{(243)^{\frac{2}{5}}}{(64)^{-\frac{1}{2}}} &= {(3^5)^{\frac{2}{5}}}{(8^2)^{-\frac{1}{2}}} \\ &= {3^{5.\frac{2}{5}}}{8^{2.\left( -\frac{1}{2} \right)}} \\ &= {3^2}{8^{-1}} \\ &= \frac{{3^2}}{8} \\ &= \frac{9}{8} \end{align}$
Jawaban: C

Soal No. 14

Diketahui $a=25$ dan $b=32$, nilai dari ${a^{\frac{1}{2}}}.{b^{-\frac{1}{5}}}$ = ….
A. $-2\frac{1}{2}$
B. $-1\frac{1}{2}$
C. $1\frac{1}{2}$
D. $2\frac{1}{2}$
E. $3\frac{1}{2}$

Penyelesaian: Lihat/Tutup

$\begin{align}{a^{\frac{1}{2}}}.{b^{-\frac{1}{5}}} &= {25^{\frac{1}{2}}}{.32^{-\frac{1}{5}}} \\ &= {{({5^2})}^{\frac{1}{2}}}.{{({2^5})}^{-\frac{1}{5}}} \\ &= {5^{2.\frac{1}{2}}}{{.2}^{5.\left( -\frac{1}{5} \right)}} \\ &= {{5.2}^{-1}} \\ &= \frac{5}{2} \\ &= 2\frac{1}{2} \end{align}$
Jawaban: D

Soal No. 15

Diketahui $a=27$ dan $b=32$. Nilai dari $({a^{\frac{2}{3}}}-{b^{\frac{2}{5}}})$ adalah ….
A. 3
B. 4
C. 5
D. 6
E. 7

Penyelesaian: Lihat/Tutup

$\begin{align}{a^{\frac{2}{3}}}-{b^{\frac{2}{5}}} &= {27^{\frac{2}{3}}}-{32^{\frac{2}{5}}} \\ &= {(3^3)^{\frac{2}{3}}}-{(2^5)^{\frac{2}{5}}} \\ &= {3^{3.\frac{2}{3}}}-{2^{5.\frac{2}{5}}} \\ &= {3^2}-{2^2} \\ &= 9-4 \\ &= 5 \end{align}$
Jawaban: C

Soal No. 16

Diketahui $a=64$ dan $b=27$. Nilai dari ${a^{\frac{1}{3}}}\times {b^{\frac{1}{3}}}$ = ….
A. $\frac{4}{3}$
B. $\frac{5}{3}$
C. $\frac{6}{3}$
D. $\frac{7}{3}$
E. $\frac{8}{3}$

Penyelesaian: Lihat/Tutup

$\begin{align}{a^{\frac{1}{3}}}\times {b^{-\frac{1}{3}}} &= {{64}^{\frac{1}{3}}}\times {27^{-\frac{1}{3}}} \\ &= {{({4^3})}^{\frac{1}{3}}}\times {(3^3)^{-\frac{1}{3}}} \\ &= {4^{3.\frac{1}{3}}}\times {3^{3.\left( -\frac{1}{3} \right)}} \\ &= 4\times {3^{-1}} \\ &= \frac{4}{3} \end{align}$
Jawaban: A

Soal No. 17

Diketahui $p=({x^{\frac{3}{2}}}+{x^{\frac{1}{2}}})({x^{\frac{1}{3}}}-{x^{-\frac{1}{3}}})$ dan $q=({x^{\frac{1}{2}}}+{x^{-\frac{1}{2}}})(x-{x^{\frac{1}{3}}})$ maka $\frac{p}{q}$ = ….
A. $\sqrt[3]{x}$
B. $\sqrt[3]{{x^2}}$
C. $x$
D. $x\sqrt[3]{x}$
E. $x\sqrt[3]{{x^2}}$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{p}{q} &= \frac{({x^{\frac{3}{2}}}+{x^{\frac{1}{2}}})({x^{\frac{1}{3}}}-{x^{-\frac{1}{3}}})}{({x^{\frac{1}{2}}}+{x^{-\frac{1}{2}}})(x-{x^{\frac{1}{3}}})} \\ &= \frac{x({x^{\frac{1}{2}}}+{x^{-\frac{1}{2}}})({x^{\frac{1}{3}}}-{x^{-\frac{1}{3}}})}{({x^{\frac{1}{2}}}+{x^{-\frac{1}{2}}}).{x^{\frac{2}{3}}}({x^{\frac{1}{3}}}-{x^{-\frac{1}{3}}})} \\ &= \frac{x}{{x^{\frac{2}{3}}}} \\ &= {x^{1-\frac{2}{3}}} \\ &= {x^{\frac{1}{3}}} \\ &= \sqrt[3]{x} \end{align}$
Jawaban: A

Soal No. 18

Bentuk $\left( \frac{{a^{-\frac{2}{3}}}}{{b^{-\frac{1}{3}}}} \right)\times {{\left( {a^{\frac{2}{3}}}.{b^{\frac{1}{2}}} \right)}^2}\div \left( \frac{{a^{\frac{1}{2}}}}{{b^{\frac{1}{3}}}} \right)$ senilai dengan ….
A. $ab$
B. $a\sqrt{b}$
C. $b\sqrt[6]{a{b^4}}$
D. $a\sqrt[6]{{b^5}}$
E. ${a^{\frac{1}{3}}}{b^{\frac{1}{2}}}$

Penyelesaian: Lihat/Tutup

$\left( \frac{{a^{-\frac{2}{3}}}}{{b^{-\frac{1}{3}}}} \right)\times {{\left( {a^{\frac{2}{3}}}.{b^{\frac{1}{2}}} \right)}^2}\div \left( \frac{{a^{\frac{1}{2}}}}{{b^{\frac{1}{3}}}} \right)$
= $\left( \frac{{a^{-\frac{2}{3}}}}{{b^{-\frac{1}{3}}}} \right)\times {{\left( {a^{\frac{2}{3}}}.{b^{\frac{1}{2}}} \right)}^2}\times \left( \frac{{b^{\frac{1}{3}}}}{{a^{\frac{1}{2}}}} \right)$
= ${a^{-\frac{2}{3}}}.{b^{\frac{1}{3}}}\times {a^{\frac{4}{3}}}.b\times {b^{\frac{1}{3}}}.{a^{-\frac{1}{2}}}$
= ${a^{-\frac{2}{3}+\frac{4}{3}-\frac{1}{2}}}.{b^{\frac{1}{3}+1+\frac{1}{3}}}$
= ${a^{\frac{1}{6}}}.{b^{\frac{5}{3}}}$
= ${a^{\frac{1}{6}}}.{b^{\frac{10}{6}}}$
= ${a^{\frac{1}{6}}}.{b^{1\frac{4}{6}}}$
= $b.{a^{\frac{1}{6}}}.{b^{\frac{4}{6}}}$
= $b.{{(a{b^4})}^{\frac{1}{6}}}$
= $b\sqrt[6]{a{b^4}}$
Jawaban: C

Soal No. 19

Bentuk sederhana dari $\frac{\sqrt[3]{{a^4}\sqrt[3]{a\sqrt{a}}}}{\sqrt{a\sqrt[3]{a}}}$ adalah ….
A. $\frac{1}{\sqrt[6]{{a^5}}}$
B. $\sqrt[6]{{a^5}}$
C. $a\sqrt[5]{a}$
D. $\frac{1}{\sqrt[6]{a}}$
E. $\sqrt[6]{a}$

Penyelesaian: Lihat/Tutup

$\begin{align}\frac{\sqrt[3]{{a^4}\sqrt[3]{a\sqrt{a}}}}{\sqrt{a\sqrt[3]{a}}} &= \frac{{{\left( {a^4}{{\left( a.{a^{\frac{1}{2}}} \right)}^{\frac{1}{3}}} \right)}^{\frac{1}{3}}}}{{{\left( a.{a^{\frac{1}{3}}} \right)}^{\frac{1}{2}}}} \\ &= \frac{{a^{\frac{4}{3}}}{{\left( a.{a^{\frac{1}{2}}} \right)}^{\frac{1}{9}}}}{{a^{\frac{1}{2}}}.{a^{\frac{1}{6}}}} \\ &= \frac{{a^{\frac{4}{3}}}.{a^{\frac{1}{9}}}.{a^{\frac{1}{18}}}}{{a^{\frac{1}{2}}}.{a^{\frac{1}{6}}}} \\ &= {a^{\frac{4}{3}+\frac{1}{9}+\frac{1}{18}-\frac{1}{2}-\frac{1}{6}}} \\ &= {a^{\frac{24}{18}+\frac{2}{18}+\frac{1}{18}-\frac{9}{18}-\frac{3}{18}}} \\ &= {a^{\frac{15}{18}}} \\ &= {a^{\frac{5}{6}}} \\ &= \sqrt[6]{{a^5}} \end{align}$
Jawaban: B

Soal No. 20

Ditentukan nilai $a=9$, $b=16$ dan $c=36$. Nilai $\sqrt{{{\left( {a^{-\frac{1}{3}}}{b^{-\frac{1}{2}}}c \right)}^3}}$ = ….
A. 3
B. 1
C. 9
D. 12
E. 18

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt{{{\left( {a^{-\frac{1}{3}}}{b^{-\frac{1}{2}}}c \right)}^3}} &= {{\left( {a^{-\frac{1}{3}}}{b^{-\frac{1}{2}}}c \right)}^{\frac{3}{2}}} \\ &= {a^{-\frac{1}{3}.\frac{3}{2}}}{b^{-\frac{1}{2}.\frac{3}{2}}}{c^{\frac{3}{2}}} \\ &= {a^{-\frac{1}{2}}}{b^{-\frac{3}{4}}}{c^{\frac{3}{2}}} \\ &= {9^{-\frac{1}{2}}}{{.16}^{-\frac{3}{4}}}{{.36}^{\frac{3}{2}}} \\ &= {{({3^2})}^{-\frac{1}{2}}}.{{({2^4})}^{-\frac{3}{4}}}.{{({6^2})}^{\frac{3}{2}}} \\ &= {3^{2.\left( -\frac{1}{2} \right)}}{{.2}^{4.\left( -\frac{3}{4} \right)}}{{.6}^{2.\frac{3}{2}}} \\ &= {3^{-1}}{{.2}^{-3}}{{.6}^3} \\ &= \frac{{6^3}}{{{3.2}^3}} \\ &= \frac{216}{24} \\ &= 9 \end{align}$
Jawaban: C

Soal No. 21

Diketahui $a=\frac{1}{8}$, $b=16$ dan $c=4$ maka nilai ${a^{-1\frac{1}{3}}}{b^{\frac{1}{4}}}{c^{-1\frac{1}{2}}}$ adalah ….
A. $\frac{1}{256}$
B. $\frac{1}{4}$
C. 1
D. 4
E. 256

Penyelesaian: Lihat/Tutup

$\begin{align}{a^{-1\frac{1}{3}}}{b^{\frac{1}{4}}}{c^{-1\frac{1}{2}}} &= {a^{-\frac{4}{3}}}{b^{\frac{1}{4}}}{c^{-\frac{3}{2}}} \\ &= {{\left( \frac{1}{8} \right)}^{-\frac{4}{3}}}.{(16)^{\frac{1}{4}}}.{{(4)}^{-\frac{3}{2}}} \\ &= {{({2^{-3}})}^{-\frac{4}{3}}}.{{({2^4})}^{\frac{1}{4}}}.{{({2^2})}^{-\frac{3}{2}}} \\ &= {2^{-3.\left( -\frac{4}{3} \right)}}{{.2}^{4.\frac{1}{4}}}{{.2}^{2.\left( -\frac{3}{2} \right)}} \\ &= {2^4}{{.2.2}^{-3}} \\ &= {2^{4+1-3}} \\ &= {2^2} \\ &= 4 \end{align}$
Jawaban: D

Soal No. 22

${{({x^2}+1)}^{\frac{2}{3}}}.{{({x^2}+1)}^{\frac{4}{3}}}$ sama dengan ….
A. ${x^4}+2{x^2}+1$
B. ${x^4}-2{x^2}+1$
C. ${x^4}+2{x^2}-1$
D. ${x^2}+2x+1$
E. ${x^2}-2x+1$

Penyelesaian: Lihat/Tutup

$\begin{align}{{({x^2}+1)}^{\frac{2}{3}}}.{{({x^2}+1)}^{\frac{4}{3}}} &= {{({x^2}+1)}^{\frac{2}{3}+\frac{4}{3}}} \\ &= {{({x^2}+1)}^2} \\ &= {x^4}+2{x^2}+1 \end{align}$
Jawaban: A

Soal No. 23

${{(\sqrt[3]{{8^2}})}^{\frac{1}{2}}}$ = ….
A. 1
B. 2
C. 3
D. 6
E. 9

Penyelesaian: Lihat/Tutup

$\begin{align}{{(\sqrt[3]{{8^2}})}^{\frac{1}{2}}} &= {{\left( {8^{\frac{2}{3}}} \right)}^{\frac{1}{2}}} \\ &= {8^{\frac{2}{3}.\frac{1}{2}}} \\ &= {8^{\frac{1}{3}}} \\ &= {(2^3)^{\frac{1}{3}}} \\ &= {2^{3.\frac{1}{3}}} \\ &= 2 \end{align}$
Jawaban: B

Soal No. 24

$\sqrt[3]{\sqrt{{2^{12}}}}$ = ….
A. 8
B. 4
C. 2
D. 1
E. 0,5

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt[3]{\sqrt{{2^{12}}}} &= \sqrt[3]{{2^{\frac{12}{2}}}} \\ &= \sqrt[3]{{2^6}} \\ &= {2^{\frac{6}{3}}} \\ &= {2^2} \\ &= 4 \end{align}$
Jawaban: B

Soal No. 25

$\sqrt[4]{\sqrt[3]{\sqrt{{9^{12}}}}}$ = ….
A. $\frac{1}{9}$
B. $\frac{1}{3}$
C. 1
D. 3
E. 9

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt[4]{\sqrt[3]{\sqrt{{9^{12}}}}} &= \sqrt[4]{\sqrt[3]{{9^{\frac{12}{2}}}}} \\ &= \sqrt[4]{\sqrt[3]{{9^6}}} \\ &= \sqrt[4]{{9^{\frac{6}{3}}}} \\ &= \sqrt[4]{{9^2}} \\ &= {9^{\frac{2}{4}}} \\ &= {{({3^2})}^{\frac{1}{2}}} \\ &= {3^{2.\frac{1}{2}}} \\ &= 3 \end{align}$
Jawaban: D

Soal No. 26

${{\left( \frac{{3^{\frac{1}{3}}}}{{9^{\frac{1}{6}}}} \right)}^{\frac{3}{5}}}$ = ….
A. 0
B. $\frac{1}{9}$
C. $\frac{1}{3}$
D. 1
E. 3

Penyelesaian: Lihat/Tutup

$\begin{align}{{\left( \frac{{3^{\frac{1}{3}}}}{{9^{\frac{1}{6}}}} \right)}^{\frac{3}{5}}} &= \frac{{3^{\frac{1}{3}.\frac{3}{5}}}}{{9^{\frac{1}{6}.\frac{3}{5}}}} \\ &= \frac{{3^{\frac{1}{5}}}}{{9^{\frac{1}{10}}}} \\ &= \frac{{3^{\frac{1}{5}}}}{{{({3^2})}^{\frac{1}{10}}}} \\ &= \frac{{3^{\frac{1}{5}}}}{{3^{2.\frac{1}{10}}}} \\ &= \frac{{3^{\frac{1}{5}}}}{{3^{\frac{1}{5}}}} \\ &= 1 \end{align}$
Jawaban: D

Soal No. 27

${{\left( \sqrt{\frac{{2^{x+2}}}{{4^{1-x}}}} \right)}^{\frac{2}{3}}}$ = ….
A. 2
B. ${2^{x}}$
C. ${2^{x+1}}$
D. ${2^{2x+1}}$
E. ${2^{2x+2}}$

Penyelesaian: Lihat/Tutup

$\begin{align}{{\left( \sqrt{\frac{{2^{x+2}}}{{4^{1-x}}}} \right)}^{\frac{2}{3}}} &= {{\left( {{\left( \frac{{2^{x+2}}}{{4^{1-x}}} \right)}^{\frac{1}{2}}} \right)}^{\frac{2}{3}}} \\ &= {{\left( \frac{{2^{x+2}}}{{2^{2(1-x)}}} \right)}^{\frac{1}{2}.\frac{2}{3}}} \\ &= {{\left( \frac{{2^{x+2}}}{{2^{2-2x}}} \right)}^{\frac{1}{3}}} \\ &= {{\left( {2^{x+2-2+2x}} \right)}^{\frac{1}{3}}} \\ &= {{\left( {2^{3x}} \right)}^{\frac{1}{3}}} \\ &= {2^{3x.\frac{1}{3}}} \\ &= {2^{x}} \end{align}$
Jawaban: B

Soal No. 28

${{\left( \frac{9{p^{1\frac{1}{3}}}{q^{\frac{3}{4}}}}{4{p^{-\frac{2}{3}}}{q^{-1\frac{1}{4}}}} \right)}^{\frac{1}{2}}}$ = ….
A. $pq$
B. $\frac{1}{2}pq$
C. $\frac{3}{2}pq$
D. $\frac{3}{2}{p^2}{q^2}$
E. $\frac{9}{4}{p^2}{q^2}$

Penyelesaian: Lihat/Tutup

$\begin{align}{{\left( \frac{9{p^{1\frac{1}{3}}}{q^{\frac{3}{4}}}}{4{p^{-\frac{2}{3}}}{q^{-1\frac{1}{4}}}} \right)}^{\frac{1}{2}}} &= {{\left( \frac{9{p^{\frac{4}{3}}}{q^{\frac{3}{4}}}}{4{p^{-\frac{2}{3}}}{q^{-\frac{5}{4}}}} \right)}^{\frac{1}{2}}} \\ &= {{\left( \frac{9}{4}{p^{\frac{4}{3}+\frac{2}{3}}}{q^{\frac{3}{4}+\frac{5}{4}}} \right)}^{\frac{1}{2}}} \\ &= {{\left( \frac{{3^2}}{{2^2}}{p^2}{q^2} \right)}^{\frac{1}{2}}} \\ &= \frac{{3^{2.\frac{1}{2}}}}{{2^{2.\frac{1}{2}}}}{p^{2.\frac{1}{2}}}{q^{2.\frac{1}{2}}} \\ &= \frac{3}{2}pq \end{align}$
Jawaban: C

Soal No. 29

$\sqrt[3]{\frac{{8^2}{x^3}}{{y^{-6}}}}$ = ….
A. $2x{y^{-2}}$
B. $4x{y^2}$
C. $4x{y^{-2}}$
D. $2x{y^2}$
E. $4xy$

Penyelesaian: Lihat/Tutup

$\begin{align}\sqrt[3]{\frac{8^2x^3}{y^{-6}}} &= \left( \frac{8^2x^3}{y^{-6}} \right)^{\frac{1}{3}} \\ &= \frac{8^{2.\frac{1}{3}}x^{3.\frac{1}{3}}}{y^{-6.\frac{1}{3}}} \\ &= \frac{8^{\frac{2}{3}}x}{y^{-2}} \\ &= (2^3)^{\frac{2}{3}}xy^2 \\ &= 2^{3.\frac{2}{3}}xy^2 \\ &= 2^2xy^2 \\ &= 4xy^2 \end{align}$
Jawaban: B

Soal No. 30

Jika $y=\sqrt[3]{(x-1)^2}+2$ maka $x$ = ….
A. $(y-2)^3+1$
B. $[(y-2)^3+1]^{\frac{1}{2}}$
C. $[(y+1)^3]^{\frac{1}{2}}+2$
D. $(y+2)^3+1$
E. $[(y-2)^3]^{\frac{1}{2}}+1$

Penyelesaian: Lihat/Tutup

$y=\sqrt[3]{(x-1)^2}+2$
$\begin{align}\sqrt[3]{(x-1)^2}+2 &= y \\ (x-1)^{\frac{2}{3}} &= y-2 \\ \left( (x-1)^{\frac{2}{3}} \right)^3 &= (y-2)^3 \\ (x-1)^{\frac{2}{3}.3} &= (y-2)^3 \\ (x-1)^2 &= (y-2)^3 \\ [(x-1)^2]^{\frac{1}{2}} &= [(y-2)^3]^{\frac{1}{2}} \\ (x-1)^{2.\frac{1}{2}} &= [(y-2)^3]^{\frac{1}{2}} \\ x-1 &= [(y-2)^3]^{\frac{1}{2}} \\ x &= [(y-2)^3]^{\frac{1}{2}}+1 \end{align}$
Jawaban: E

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