Kumpulan Artikel Ilmiah: Pembahasan Soal Latihan Eksponen

Nilai x yang memenuhi persamaan $\frac{\sqrt[3]{\frac{1}{9^{2-x}}}}{27}$ = 3^x+1 adalah …

A. -16

B. -7

C. 4

D. 5

E. 6

PEMBAHASAN :

$\frac{\sqrt[3]{\frac{1}{9^{2-x}}}}{27}$ = 3^x+1

$(\frac{1}{9^{2-x}})^{1/3}$ = 3^x+1.27

9^{-1/3(2 – x)} = 3^{x + 1}.3³

3^{2(-2/3 + x/3)} = 3^{x + 4}

2( $-\frac{2}{3}$ + $\frac{x}{3}$ ) = x + 4 [kali 3 kedua ruas]

-4 + 2x = 3x + 12

-16 = x

JAWABAN : A
Nilai x yang memenuhi persamaan $\frac{0,09^{1/2(x-3)}}{0,3^{3x+1}}$ = 1 adalah …

A. -2

B. -1

C. 0

D. 1

E. 2

PEMBAHASAN :

$\frac{0,09^{1/2(x-3)}}{0,3^{3x+1}}$ = 1

0,3^2(1/2(x-3)) = 0,3^3x+1

2( $\frac{1}{2}$ (x-3)) = 3x + 1

x – 3 = 3x + 1

-4 = 2x

-2 = x

JAWABAN : A
Diberikan persamaan $(\sqrt[3]{\frac{1}{243}})^{3x}$ = $(\frac{3}{3^{x-2}})^2$ $\sqrt[3]{\frac{1}{9}}$ , hitunglah nilai dari 1 – $\frac{3}{4}$ x = …

A. $1\frac{3}{16}$

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

C. $1\frac{3}{4}$

D. $2\frac{1}{3}$

E. $2\frac{3}{4}$

PEMBAHASAN :

$(\sqrt[3]{\frac{1}{243}})^{3x}$ = $(\frac{3}{3^{x-2}})^2$ $\sqrt[3]{\frac{1}{9}}$

(3^-5/3)^3x = (3^1-(x-2))² 3^-2/3

3^-5x = (3^3-x)² 3^-2/3

3^-5x = 3^6-2x 3^-2/3

3^-5x = 3^{6 – 2x – 2/3}

-5x = 6 – 2x – $\frac{2}{3}$

-3x = $\frac{16}{3}$

x = $-\frac{16}{9}$

sehingga 1 – $\frac{3}{4}$ x = 1 – $\frac{3}{4}$ $-\frac{16}{9}$

= $\frac{36}{36}$ + $\frac{48}{36}$

= $\frac{84}{36}$

= $2\frac{1}{3}$

JAWABAN : D
$(\frac{a^{2/3}}{b^{1/2}})^{-1}$ . $(a^{2/3}b^{1/2})$ : $\frac{b^{1/2}}{a^{4/3}}$ = …

A. $\sqrt{a.b}$

B. $\sqrt{a}$ .b

C. a.b

D. $a\sqrt{b}$

E. a^1/3.b¹²

PEMBAHASAN :

$(\frac{a^{2/3}}{b^{1/2}})^{-1}$ . $(a^{2/3}b^{1/2})$ : $\frac{b^{1/2}}{a^{4/3}}$ = $\frac{a^{-2/3}.b^{1/2}.a^{2/3}.a^{4/3}}{b^{-1/2}.b^{1/2}}$

= $\frac{a^{-2/3+2/3+4/2}.b^{1/2}}{b^{1/2-1/2}}$

= a^4/3 . b^1/2

JAWABAN : C
Jika a^3/2 = b^-3/2 c^3/4, maka c dinyatakan dalam a dan b adalah …

A. $\frac{4}{3}$ a^1/2 b^3/2

B. $\frac{4}{3}$ a^1/2 b^-3/2

C. a^1/2 b^3/2

D. a^2/3 b^-2

E. a² b²

PEMBAHASAN :

a^3/2 / b^-3/2 = c^3/4

a^3/2 . b^3/2 = c^3/4

(a^3/2 . b^3/2)^4/3 = c

a² . b² = c

JAWABAN : E
Penyelesaian persamaan $\sqrt{3^{2x+1}}$ = $9^{x-2}$

A. 0

B. $1\frac{1}{2}$

C. 2

D. $3\frac{1}{2}$

E. $4\frac{1}{2}$

PEMBAHASAN :

$\sqrt{3^{2x+1}}$ = $9^{x-2}$

(3^{2x + 1})^1/2 = 3^{2(x – 2)}

3^{x + 1/2} = 3^{2x – 4}

x + $\frac{1}{2}$ = 2x – 4

x = $\frac{9}{2}$

JAWABAN : E
Nilai x yang memenuhi persamaan $\frac{1}{27^{3x-7}}$ = $\sqrt{3^{2-2x}}$ adalah …

A. $-\frac{5}{4}$

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

C. 1

D. 2

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

PEMBAHASAN :

$\frac{1}{27^{3x-7}}$ = $\sqrt{3^{2-2x}}$

27^{-(3x – 7)} = (3^2-2x)^1/2

3^{-3(3x – 7)} = 3^{1 – x}

-3(3x – 7) = 1 – x

-9x + 21 = 1 – x

8x = 20

x = $\frac{5}{2}$

JAWABAN : E
Diketahui x^1/2 + x^-1/2 = 3, maka nilai x + x^-1 = …

A. 7

B. 8

C. 9

D. 10

E. 11

PEMBAHASAN :

(x^1/2 + x^-1/2)² = x + 2. x^1/2. ^x-1/2 + x^-1

3² = x + 2 + x^-1

7 = x + x^-1

JAWABAN : A
Jika x memenuhi persamaan 3x^0,4 – 9( $\frac{1}{3}$ )^0,6 = 0, maka 3x – x² sama dengan …

A. 3^0,4

B. 3^0,5

C. 3^-0,25

D. $\frac{6}{9}$

E. 0

PEMBAHASAN :

3x^0,4 – 3²(3^-1)^0,6 = 0

3x^0,4 – 3².3^-0,6 = 0

3x^0,4 – 3^1,4 = 0

3x^0,4 = 3^1,4

3. x^0,4 = 3.3^0,4

x^0,4 = 3^0,4

x = 3

sehingga 3x – x² = 3.3 – 3² = 0

JAWABAN : E
Jika n bilangan bulat, maka nilai dari $\frac{2^{n+2}.6^{n-4}}{12^{n-1}}$ = …

A. $\frac{1}{27}$

B. $\frac{1}{16}$

C. $\frac{1}{9}$

D. $\frac{1}{8}$

E. $\frac{1}{3}$

PEMBAHASAN :

$\frac{2^{n+2}.6^{n-4}}{12^{n-1}}$ = $\frac{2^{n+2}.6^{n-4}}{(2.6)^{n-1}}$

= $\frac{2^{n+2}.6^{n-4}}{2^{n-1}.6^{n-1}}$

= 2^{n + 2} . 2^{-(n – 1)} . 6^{n – 4} . 6^{-(n – 1)}

= 2^{n + 2} . 2^{-n + 1} . 6^{n – 4} . 6^{-n + 1}

= 2³ . 6^-3

= $\frac{2.2.2}{6.6.6}$

= $\frac{1}{3.3.3}$

= $\frac{1}{27}$

JAWABAN : A