HOTS Zone : Pangkat (Eksponen)

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StandarSNBTHOTS

12

No. 1

Jika
2^a = 3,
3^b = 4,
4^c = 5,
5^d = 6,
6^e = 7,
7^f = 8, dan
8^g = 9,
maka hasil dari bcdef(a + g) adalah....

OMVN 2018

ALTERNATIF PENYELESAIAN

CARA 1

$\begin{array}{rl}{2}^{abcdef}& ={\left(2^a\right)}^{bcdef}\\ & =3^{bcdef}\\ & ={\left(3^b\right)}^{cdef}\\ & =4^{cdef}\\ & ={\left(4^c\right)}^{def}\\ & =5^{def}\\ & ={\left(5^d\right)}^{ef}\\ & =6^{ef}\\ & ={\left(6^e\right)}^f\\ & =7^f\\ & =8\\ & =2^3\\ abcdef& =3\end{array}$

$\begin{array}{rl}{3}^{bcdefg}& ={\left(3^{bcdef}\right)}^g\\ & =8^g\\ & =9\\ & =3^2\\ bcdefg& =2\end{array}$

$\begin{array}{rl}bcdfg(a+g)& =abcdef+bcdefg\\ & =3+2\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black5}}}}\end{array}$

CARA 2

$2^a=3\Rightarrow a={}^2\log 3$
$3^b=4\Rightarrow b={}^3\log 4$
$4^c=5\Rightarrow c={}^4\log 5$
$5^d=6\Rightarrow d={}^5\log 6$
$6^e=7\Rightarrow e={}^6\log 7$
$7^f=8\Rightarrow f={}^7\log 8$
$8^g=9\Rightarrow g={}^8\log 9$

$\begin{array}{rl}bcdef(a+g)& ={}^3\log 4\cdot {}^4\log 5\cdot {}^5\log 6\cdot {}^6\log 7\cdot {}^7\log 8({}^2\log 3+{}^8\log 9)\\ & ={}^3\log 8({}^2\log 3+{}^8\log 9)\\ & ={}^2\log 3\cdot {}^3\log 8+{}^3\log 8\cdot {}^8\log 9\\ & ={}^2\log 8+{}^3\log 9\\ & =3+2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$

Jadi, bcdef(a + g) = 5.

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No. 2

Diketahui A = {0, 1, 2, 3, 4}; a, b, c adalah tiga anggota yang berbeda dari A, dan (a^b)^c = n. Nilai maksimum dari n adalah

1. 4096
2. 6561

3. 9561
4. 9651

OSK SMP 2019

ALTERNATIF PENYELESAIAN

Jika a = 0 maka n = 0
Jika a = 1 maka n = 1
Jadi kemungkinan nilai untuk a adalah 2, 3, atau 4.
Jika a = 2, maka:
n = (2³)⁴ = 2¹² = 4096

Jika a = 3, maka:
n = (3²)⁴ = 3⁸ = 6561

Jika a = 4, maka:
n = (4²)³ = 4⁶ = 4096

Jadi, nilai maksimum dari n adalah 6561.
JAWAB: B

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No. 3

Diketahui m^{x + 1} = n^{2 − x} = p. Jika x₁ dan x₂ memenuhi persamaan m × n = p⁶, tentukan nilai x₁² + x₂².

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{m}^{x+1}& =p\\ m& =p^{\frac{1}{x+1}}\end{array}$

$\begin{array}{rl}{n}^{2-x}& =p\\ n& =p^{\frac{1}{2-x}}\end{array}$

$\begin{array}{rl}m\times n& =p^6\\ p^{\frac{1}{x+1}}\times p^{\frac{1}{2-x}}& =p^6\\ p^{\frac{1}{x+1}+\frac{1}{2-x}}& =p^6\\ p^{\frac{2-x+x+1}{(x+1)(2-x)}}& =p^6\\ p^{\frac{3}{-x^2+x+2}}& =p^6\\ \frac{3}{-x^2+x+2}& =6\\ 3& =-6x^2+6x+12\\ 6x^2-6x-9& =0\\ 2x^2-2x-3& =0\end{array}$

$$\begin{gathered} x_1+x_2=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{-2}{2}}=1 \\ {x}_1{x}_2={\displaystyle \frac{c}{a}}={\displaystyle \frac{-3}{2}}=-{\displaystyle \frac{3}{2}} \end{gathered}$$

$\begin{array}{rl}{x_1}^2+{x_2}^2& ={(x_1+x_2)}^2-2x_1{x}_2\\ & =(1{)}^2-2(-{\displaystyle \frac{3}{2}})\\ & =1+3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

Jadi, x₁² + x₂² = 4.

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No. 4

Jika a dan b adalah bilangan bulat positif yang memenuhi a^b = 2³⁰ − 2²⁹, maka nilai a + b adalah

1. 29
2. 30
3. 31

4. 32
5. 33

Matematika Idhamdaz

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{a}^b& =2^{30}-2^{29}\\ & =2^{29}(2-1)\\ & =2^{29}\end{array}$

a = 2
b = 29

a + b = 2 + 29 = 31

Jadi, a + b = 31.

Related: loading

JAWAB: C

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No. 5

Jika 3^x = 4^y = 12^z buktikanlah bahwa $z={\displaystyle \frac{xy}{x+y}}$

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ALTERNATIF PENYELESAIAN

3^x = 4^y = 12^z = k
$3=k^{\frac{1}{x}}$
$4=k^{\frac{1}{y}}$
$12=k^{\frac{1}{z}}$

$\begin{array}{rl}12& =3\cdot 4\\ k^{\frac{1}{z}}& =k^{\frac{1}{x}}{k}^{\frac{1}{y}}\\ k^{\frac{1}{z}}& =k^{\frac{1}{x}+\frac{1}{y}}\\ \frac{1}{z}& =\frac{1}{x}+\frac{1}{y}\\ & ={\displaystyle \frac{x+y}{xy}}\\ z& ={\displaystyle \frac{xy}{x+y}}\end{array}$

Jadi, terbukti bahwa $z={\displaystyle \frac{xy}{x+y}}$.

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No. 6

Jika $x={(1+{\displaystyle \frac{1}{n}})}^n$ dan $y={(1+{\displaystyle \frac{1}{n}})}^{n+1}$, buktikan bahwa y^x = x^y

Canadian MO 1974

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{y}^x& =x^y\\ {\left({(1+{\displaystyle \frac{1}{n}})}^{n+1}\right)}^x& ={\left({(1+{\displaystyle \frac{1}{n}})}^n\right)}^y\\ {(1+{\displaystyle \frac{1}{n}})}^{(n+1)x}& ={(1+{\displaystyle \frac{1}{n}})}^{ny}\\ (n+1)x& =ny\end{array}$
Perhatikan bahwa membuktikan y^x = x^y sama saja dengan membuktikan (n + 1)x = ny.

$\begin{array}{rl}y& ={(1+{\displaystyle \frac{1}{n}})}^{n+1}\\ & ={(1+{\displaystyle \frac{1}{n}})}^n\cdot (1+{\displaystyle \frac{1}{n}})\\ & =x\left({\displaystyle \frac{n+1}{n}}\right)\\ ny& =(n+1)x\end{array}$
TERBUKTI

Jadi, terbukti bahwa y^x = x^y.

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No. 7

Jika a = 60(99)⁹⁹ + 40(99)⁹⁹, b = 99¹⁰⁰, dan c = 90(99)⁹⁹, maka hubungan antara a, b, c yang paling tepat adalah ....

1. c < a < b
2. b < c < a
3. a < b < c

4. c < b < a
5. b < a < c

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}a& =60(99{)}^{99}+40(99{)}^{99}\\ & =100(99{)}^{99}\end{array}$

$\begin{array}{rl}b& ={99}^{100}\\ & ={99}^{1+99}\\ & =99(99{)}^{99}\end{array}$

$\begin{array}{rcccl}90(99{)}^{99}& <& 99(99{)}^{99}& <& 100(99{)}^{99}\\ c& <& b& <& a\end{array}$

Jadi, hubungan antara a, b, c yang paling tepat adalah c < b < a.
JAWAB: D

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No. 8

Diberikan bilangan bulat positif empat angka A dan B sedemikian sehingga A × B = 16⁵ + 2¹⁰. Hasil dari A + B = ....

1. 2045
2. 2046
3. 2047

4. 2048
5. 2049

C S C POSI 2021

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}A\times B& ={16}^5+2^{10}\\ & ={\left(2^4\right)}^5+2^{10}\\ & =2^{20}+2^{10}\\ & =2^{10}(2^{10}+1)\\ & =1024(1025)\end{array}$

$\begin{array}{rl}A+B& =1024+1025\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2049}}}}\end{array}$

Jadi, A + B = 2049.
JAWAB: E

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No. 9

Bentuk paling sederhana dari ${\displaystyle \frac{3^{2018}-2\cdot 3^{2015}+125}{3^{2017}-4\cdot 3^{2015}+25}}$ adalah ....

1. 1
2. 5
3. 10

4. 15
5. 25

Forum Matematika

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{3^{2018}-2\cdot 3^{2015}+125}{3^{2017}-4\cdot 3^{2015}+25}}& ={\displaystyle \frac{3^{2015}(3^3-2)+125}{3^{2015}(3^2-4)+25}}\\ & ={\displaystyle \frac{3^{2015}\left(25\right)+125}{3^{2015}\left(5\right)+25}}\\ & ={\displaystyle \frac{25(3^{2015}+5)}{5(3^{2015}+5)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$$

Jadi, bentuk paling sederhana dari ${\displaystyle \frac{3^{2018}-2\cdot 3^{2015}+125}{3^{2017}-4\cdot 3^{2015}+25}}$ adalah 5.
JAWAB: B

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No. 10

Di antara bilangan-bilangan berikut ini, yang memiliki nilai terbesar adalah ....

1. 2⁷⁷
2. 3⁶⁶
3. 4⁵⁵

4. 5⁴⁴
5. 6³³

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ALTERNATIF PENYELESAIAN

2⁷⁷
(2⁷)¹¹
125¹¹

3⁶⁶
(3⁶)¹¹
729¹¹

4⁵⁵
(4⁵)¹¹
1024¹¹

5⁴⁴
(5⁴)¹¹
625¹¹

6³³
(6³)¹¹
216¹¹

Jadi, yang memiliki nilai terbesar adalah 4⁵⁵.
JAWAB: C

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