SBMPTN Zone : Persamaan Eksponen

Berikut ini adalah kumpulan soal mengenai Persamaan Eksponen tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

---

No. 1

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. $^6\negmedspace\log7$

Ganesha Operation

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

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

---

No. 2

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. $3+2\ {^3\negthinspace\log4}$

SUMBER

Jadi, $x=1+3\ {^3\negthinspace\log6}$.
JAWAB: A

---

No. 3

$\dfrac{8^x+27^x}{12^x+18^x}=\dfrac76$, jumlah dari nilai semua penyelesaian $x$ yang mungkin adalah....

1. $-2$
2. $-1$
3. $0$

4. $1$
5. $2$

SUMBER

$\begin{aligned}\dfrac{8^x+27^x}{12^x+18^x}&=\dfrac76\\[8pt]\dfrac{\left(2^3\right)^x+\left(3^3\right)^x}{(6\cdot2)^x+(6\cdot3)^x}&=\dfrac76\\[8pt]\dfrac{2^{3x}+3^{3x}}{6^x\cdot2^x+6^x\cdot3^x}&=\dfrac76\\[8pt]\dfrac{\left(2^x\right)^3+\left(3^x\right)^3}{6^x\left(2^x+3^x\right)}&=\dfrac76\\[8pt]\dfrac{\left(2^x+3^x\right)\left(\left(2^x\right)^2-\left(2^x\right)\left(3^x\right)+\left(3^x\right)^2\right)}{(2\cdot3)^x\left(2^x+3^x\right)}&=\dfrac76\\[8pt]\dfrac{\left(2^x\right)^2-\left(2^x\right)\left(3^x\right)+\left(3^x\right)^2}{\left(2^x\right)\left(3^x\right)}&=\dfrac76\\[8pt]\dfrac{\left(2^x\right)^2}{\left(2^x\right)\left(3^x\right)}-\dfrac{\left(2^x\right)\left(3^x\right)}{\left(2^x\right)\left(3^x\right)}+\dfrac{\left(3^x\right)^2}{\left(2^x\right)\left(3^x\right)}&=\dfrac76\\[8pt]\dfrac{2^x}{3^x}-1+\dfrac{3^x}{2^x}&=\dfrac76\\[8pt]\left(\dfrac23\right)^x+\left(\dfrac32\right)^x&=\dfrac{13}6\end{aligned}$

Misal $\left(\dfrac23\right)^x=y$

$\begin{aligned}y+\dfrac1y&=\dfrac{13}6\\[8pt]y^2+1&=\dfrac{13}6y\\[8pt]y^2-\dfrac{13}6y+1&=0\end{aligned}$

$\begin{aligned}y_1y_2&=\dfrac{c}a\\[8pt]\left(\dfrac23\right)^{x_1}\left(\dfrac23\right)^{x_2}&=\dfrac11\\[8pt]\left(\dfrac23\right)^{x_1+x_2}&=1\\[8pt]x_1+x_2&=0\end{aligned}$

Jadi, jumlah dari nilai semua penyelesaian $x$ yang mungkin adalah 0.
JAWAB: C

---

No. 4

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. $16$

$\begin{aligned}4^{\frac{x}2}-5\cdot2^{\frac{x}2+1}-4\cdot2^{\frac{x}2}+a&=0\\\left(2^{\frac{x}2}\right)^2+(\cdots)2^{\frac{x}2}+a&=0\end{aligned}$

Jadi, $a=10$.
JAWAB: A

---

No. 5

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

$\begin{aligned}9^x-3^{x+1}-3^{x+2}-3\cdot3^{x+3}+a&=0\\\left(3^2\right)^x-3^x\cdot3^1-3^x\cdot3^2-3\cdot3^x\cdot3^3+a&=0\\3^{2x}-3\cdot3^x-9\cdot3^x-3\cdot3^x\cdot27+a&=0\\\left(3^x\right)^2-3\cdot3^x-9\cdot3^x-81\cdot3^x+a&=0\\\left(3^x\right)^2-93\cdot3^x+a&=0\end{aligned}$
$A=1$, $B=-93$, $C=a$

CARA 1	CARA 2

Jadi, $a=8$.
JAWAB: D

Share this post