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
- 5\left({^6\negmedspace\log7}\right)
- 4\left({^6\negmedspace\log7}\right)
- 3\left({^6\negmedspace\log7}\right)
- 2\left({^6\negmedspace\log7}\right)
- ^6\negmedspace\log7
\(\eqalign{
\sqrt[3]{6^x+6^{x+1}}&=7\\
6^x+6\cdot6^x&=7^3\\
7\cdot6^x&=7^3\\
6^x&=7^2\\
x&={^6\negmedspace\log7^2}\\
&=\boxed{\boxed{2\left({^6\negmedspace\log7}\right)}}
}\)
No. 2
Diketahui
6^{x+2} = 18^{x-1}, maka nilai
x adalah . . .
- 1+3\ {^3\negthinspace\log6}
- 1+4\ {^3\negthinspace\log6}
- 2+3\ {^3\negthinspace\log4}
- 2+4\ {^3\negthinspace\log4}
- 3+2\ {^3\negthinspace\log4}
\(\begin{aligned}
6^{x+2}&= 18^{x-1}\\
6^{x-1+3}&= 18^{x-1}\\
6^{x-1}\cdot6^3&= 18^{x-1}\\
6^3&=\dfrac{18^{x-1}}{6^{x-1}}\\
6^3&=3^{x-1}\\
x-1&={^3\negthinspace\log6^3}\\
x&=1+3\ {^3\negthinspace\log6}
\end{aligned}\)
No. 3
\dfrac{8^x+27^x}{12^x+18^x}=\dfrac76, jumlah dari nilai semua penyelesaian
x yang mungkin adalah....
\(\begin{aligned}
\dfrac{8^x+27^x}{12^x+18^x}&=\dfrac76\\[8pt]
\dfrac{\left(2^3\right)^x+\left(3^3\right)^x}{(6\cdot2)^x+(6\cdot3)^x}&=\dfrac76\\[8pt]
\dfrac{2^{3x}+3^{3x}}{6^x\cdot2^x+6^x\cdot3^x}&=\dfrac76\\[8pt]
\dfrac{\left(2^x\right)^3+\left(3^x\right)^3}{6^x\left(2^x+3^x\right)}&=\dfrac76\\[8pt]
\dfrac{\left(2^x+3^x\right)\left(\left(2^x\right)^2-\left(2^x\right)\left(3^x\right)+\left(3^x\right)^2\right)}{(2\cdot3)^x\left(2^x+3^x\right)}&=\dfrac76\\[8pt]
\dfrac{\left(2^x\right)^2-\left(2^x\right)\left(3^x\right)+\left(3^x\right)^2}{\left(2^x\right)\left(3^x\right)}&=\dfrac76\\[8pt]
\dfrac{\left(2^x\right)^2}{\left(2^x\right)\left(3^x\right)}-\dfrac{\left(2^x\right)\left(3^x\right)}{\left(2^x\right)\left(3^x\right)}+\dfrac{\left(3^x\right)^2}{\left(2^x\right)\left(3^x\right)}&=\dfrac76\\[8pt]
\dfrac{2^x}{3^x}-1+\dfrac{3^x}{2^x}&=\dfrac76\\[8pt]
\left(\dfrac23\right)^x+\left(\dfrac32\right)^x&=\dfrac{13}6
\end{aligned}\)
Misal \left(\dfrac23\right)^x=y
\(\begin{aligned}
y+\dfrac1y&=\dfrac{13}6\\[8pt]
y^2+1&=\dfrac{13}6y\\[8pt]
y^2-\dfrac{13}6y+1&=0
\end{aligned}\)
\(\begin{aligned}
y_1y_2&=\dfrac{c}a\\[8pt]
\left(\dfrac23\right)^{x_1}\left(\dfrac23\right)^{x_2}&=\dfrac11\\[8pt]
\left(\dfrac23\right)^{x_1+x_2}&=1\\[8pt]
x_1+x_2&=0
\end{aligned}\)
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=}
\(\begin{aligned}
4^{\frac{x}2}-5\cdot2^{\frac{x}2+1}-4\cdot2^{\frac{x}2}+a&=0\\
\left(2^{\frac{x}2}\right)^2+(\cdots)2^{\frac{x}2}+a&=0
\end{aligned}\)
\(\begin{aligned}
x_1+x_2&={^2\negmedspace\log}5^2+2\\
2^{x_1+x_2}&=2^{^2\negmedspace\log5^2+2}\\
\left(2^{\frac{x_1+x_2}2}\right)^2&=2^{^2\negmedspace\log5^2}\cdot2^2\\
\left(2^{\frac{x_1}2}\cdot2^{\frac{x_2}2}\right)^2&=5^2\cdot2^2\\
\left(\dfrac{a}1\right)^2&=(5\cdot2)^2\\
a^2&=10^2\\
a&=\boxed{\boxed{10}}
\end{aligned}\)
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= ....
\(\begin{aligned}
9^x-3^{x+1}-3^{x+2}-3\cdot3^{x+3}+a&=0\\
\left(3^2\right)^x-3^x\cdot3^1-3^x\cdot3^2-3\cdot3^x\cdot3^3+a&=0\\
3^{2x}-3\cdot3^x-9\cdot3^x-3\cdot3^x\cdot27+a&=0\\
\left(3^x\right)^2-3\cdot3^x-9\cdot3^x-81\cdot3^x+a&=0\\
\left(3^x\right)^2-93\cdot3^x+a&=0
\end{aligned}\)
A=1,
B=-93,
C=a
| CARA 1 | CARA 2 |
|---|
| \(\begin{aligned}
3^{x_1+x_2}&=3^{3\cdot{^3\negmedspace\log2}}\\
3^{x_1}\cdot3^{x_2}&=3^{{^3\negmedspace\log}2^3}\\
\dfrac{C}A&=3^{{^3\negmedspace\log}8}\\
\dfrac{a}1&=8\\
a&=8
\end{aligned}\) | \(\begin{aligned}
x_1+x_2&=3\cdot{^3\negmedspace\log}2\\
{^3\negmedspace\log}\dfrac{C}A&={^3\negmedspace\log}2^3\\
{^3\negmedspace\log}\dfrac{a}1&={^3\negmedspace\log}8\\
{^3\negmedspace\log}a&={^3\negmedspace\log}8\\
a&=8
\end{aligned}\) |
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