SBMPTN Zone : Persamaan Eksponen (Exponential Equation)

Berikut ini adalah kumpulan soal mengenai persamaan eksponen tipe SBMPTN. Jika ingin bertanya soal, silahkan tulis soalnya di kolom komentar. Kamu juga bisa bertanya di grup Telegram https://t.me/matematikazoneidgrup atau grup Facebook https://web.facebook.com/groups/matematikazoneid/.

Tipe:

Standar SBMPTN HOTS

No. 1

Jika x_1 dan x_2 adalah akar-akar {4^{\frac{x}2}-5\cdot2^{\frac{x}2+1}-4\cdot2^{\frac{x}2}+a=0} dan {x_1+x_2={^2\negmedspace\log}5^2+2}, maka {a=}

1. 10
2. 5
   
3. 25
   

4. 4
5. 16
   

$\begin{array}{rl}{4}^{\frac{x}{2}}-5\cdot 2^{\frac{x}{2}+1}-4\cdot 2^{\frac{x}{2}}+a& =0\\ {\left(2^{\frac{x}{2}}\right)}^2+(\cdots )2^{\frac{x}{2}}+a& =0\end{array}$

$\begin{array}{rl}{x}_1+x_2& ={}^2\log 5^2+2\\ 2^{x_1+x_2}& =2^{{}^2\log 5^2+2}\\ {\left(2^{\frac{x_1+x_2}{2}}\right)}^2& =2^{{}^2\log 5^2}\cdot 2^2\\ {(2^{\frac{x_1}{2}}\cdot 2^{\frac{x_2}{2}})}^2& =5^2\cdot 2^2\\ {\left({\displaystyle \frac{a}{1}}\right)}^2& =(5\cdot 2{)}^2\\ a^2& ={10}^2\\ a& =\boxed{{\displaystyle \boxed{{\displaystyle 10}}}}\end{array}$

Jadi, a=10.
JAWAB: A

No. 2

Jika x_1, x_2 adalah akar-akar {9^x-3^{x+1}-3^{x+2}-3\cdot3^{x+3}+a=0} dimana {x_1+x_2=3\cdot{^3\negmedspace\log}2}, maka a= ....

1. 27
2. 16
   
3. 9
   

4. 8
5. 4
   

RUANG GURU

$\begin{array}{rl}{9}^x-3^{x+1}-3^{x+2}-3\cdot 3^{x+3}+a& =0\\ {\left(3^2\right)}^x-3^x\cdot 3^1-3^x\cdot 3^2-3\cdot 3^x\cdot 3^3+a& =0\\ 3^{2x}-3\cdot 3^x-9\cdot 3^x-3\cdot 3^x\cdot 27+a& =0\\ {\left(3^x\right)}^2-3\cdot 3^x-9\cdot 3^x-81\cdot 3^x+a& =0\\ {\left(3^x\right)}^2-93\cdot 3^x+a& =0\end{array}$
A=1, B=-93, C=a

CARA 1	CARA 2
$\begin{array}{rl}{3}^{x_1+x_2}& =3^{3\cdot {}^3\log 2}\\ 3^{x_1}\cdot 3^{x_2}& =3^{{}^3\log 2^3}\\ {\displaystyle \frac{C}{A}}& =3^{{}^3\log 8}\\ {\displaystyle \frac{a}{1}}& =8\\ a& =8\end{array}$	$\begin{array}{rl}{x}_1+x_2& =3\cdot {}^3\log 2\\ {}^3\log {\displaystyle \frac{C}{A}}& ={}^3\log 2^3\\ {}^3\log {\displaystyle \frac{a}{1}}& ={}^3\log 8\\ {}^3\log a& ={}^3\log 8\\ a& =8\end{array}$

Jadi, a=8.
JAWAB: D

No. 3

Diketahui 6^{x+2} = 18^{x-1}, maka nilai x adalah . . .

1. {1+3\ {^3\negthinspace\log6}}
2. {1+4\ {^3\negthinspace\log6}}
3. {2+3\ {^3\negthinspace\log4}}

4. {2+4\ {^3\negthinspace\log4}}
5. 3+2\ {^3\negthinspace\log4}

learncy

$\begin{array}{rl}{6}^{x+2}& ={18}^{x-1}\\ 6^{x-1+3}& ={18}^{x-1}\\ 6^{x-1}\cdot 6^3& ={18}^{x-1}\\ 6^3& ={\displaystyle \frac{{18}^{x-1}}{6^{x-1}}}\\ 6^3& =3^{x-1}\\ x-1& ={}^3\log 6^3\\ x& =1+3{}^3\log 6\end{array}$

Jadi, x=1+3\ {^3\negthinspace\log6}.

No. 4

Nilai c yang memenuhi {(0{,}11)^{x^2+3x-c}\gt(0{,}0121)^{x^2+2x+3}} untuk semua x bilangan riil adalah

1. c\gt-\dfrac{23}4
2. c\gt-6
3. c\lt-\dfrac{23}4

4. c\lt-6
5. c\lt6

SKMP September 2020

$\begin{array}{rl}(0,11{)}^{x^2+3x-c}& >(\mathrm{0,012}1{)}^{x^2+2x+3}\\ (0,11{)}^{x^2+3x-c}& >{\left(0,{11}^2\right)}^{x^2+2x+3}\\ (0,11{)}^{x^2+3x-c}& >{\left(0,11\right)}^{2x^2+4x+6}\\ x^2+3x-c& <2x^2+4x+6\\ x^2+x+c+6& >0\end{array}$

$\begin{array}{rl}D& <0\\ b^2-4ac& <0\\ 1^2-4(1)(c+6)& <0\\ 1-4c-24& <0\\ -4c-23& <0\\ -4c& <23\\ c& >-{\displaystyle \frac{23}{4}}\end{array}$

Jadi, c\gt-\dfrac{23}4.
JAWAB: A

No. 5

Diketahui {f(x) = 2^{5- x} + 2x - 12}. Jika {f\left(x_1\right) =f\left(x_2\right) = 0} maka {x_1x_2 =}

1. -6
2. -5
3. 4

4. 5
5. 6

SKMP Juli 2020

$\begin{array}{rl}f(x)& =0\\ 2^{5-x}+2x-12& =0\\ {\displaystyle \frac{2^5}{2^x}}+2x-12& =0\\ {\displaystyle \frac{32}{2^x}}+2x-12& =0& \times 2^x\\ 32+{\left(2^x\right)}^2-12\cdot 2^x& =0\\ {\left(2^x\right)}^2-12\cdot 2^x+32& =0\\ (2^x-4)(2^x-8)& =0\end{array}$

---

$\begin{array}{rl}{2}^{x_1}-4& =0\\ 2^{x_1}& =4\\ 2^{x_1}& =2^2\\ x_1& =2\end{array}$

$\begin{array}{rl}{2}^{x_2}-8& =0\\ 2^{x_2}& =8\\ 2^{x_2}& =2^3\\ x_2& =3\end{array}$

---

{x_1x_2=(2)(3)=\boxed{\boxed{6}}}

Jadi, {x_1x_2 =6}.
JAWAB: E

No. 6

Nilai x yang memenuhi dari persamaan {\sqrt[3]{6^x+6^{x+1}}=7} adalah

1. 5\left({^6\negmedspace\log7}\right)
2. 4\left({^6\negmedspace\log7}\right)
3. 3\left({^6\negmedspace\log7}\right)

4. 2\left({^6\negmedspace\log7}\right)
5. ^6\negmedspace\log7

SKMP Juli 2020

$\begin{array}{rl}\sqrt[3]{6^x+6^{x+1}}& =7\\ 6^x+6\cdot 6^x& =7^3\\ 7\cdot 6^x& =7^3\\ 6^x& =7^2\\ x& ={}^6\log 7^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2\left({}^6\log 7\right)}}}}\end{array}$

Jadi, x=2\left({^6\negmedspace\log7}\right).
JAWAB: D