HOTS Zone : Persamaan Eksponen
Table of Contents
Tipe:
No.
\(\dfrac{8^x+27^x}{12^x+18^x}=\dfrac76\), jumlah dari nilai semua penyelesaian x yang mungkin adalah....- −2
- −1
- 0
- 1
- 2
ALTERNATIF PENYELESAIAN
\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
JAWAB: C
No.
Diketahui sistem persamaan:\(3^{\frac{a}b}=4\)
Nilai \(\dfrac1a+\dfrac1b=\) ....
ALTERNATIF PENYELESAIAN
\(\eqalign{
3^{\frac{a}b}&=4\\
\left(3^{\frac{a}b}\right)^b&=4^b\\
3^a&=4^b
}\)
\(\eqalign{ 3^a+4^b&=6\\ 3^a+3^a&=6\\ 2\cdot3^a&=6\\ 3^a&=3\\ a&=1\\ \dfrac1a&=1 }\)
\(\eqalign{ 3^a+4^b&=6\\ 3^a+3^a&=6\\ 2\cdot3^a&=6\\ 3^a&=3\\ a&=1\\ \dfrac1a&=1 }\)
\(\eqalign{
3^{\frac{a}b}&=4\\
3^{\frac1b}&=4\\
\dfrac1b&={^3\negmedspace\log4}
}\)
\(\dfrac1a+\dfrac1b=\boxed{\boxed{1+{^3\negmedspace\log4}}}\)
\(\dfrac1a+\dfrac1b=\boxed{\boxed{1+{^3\negmedspace\log4}}}\)
Jadi, \(\dfrac1a+\dfrac1b=1+{^3\negmedspace\log4}\).
No.
JikaALTERNATIF PENYELESAIAN
\(30=k^{\frac1x}\), \(2=k^{-\frac1a}\), \(3=k^{-\frac1b}\), \(5=k^{-\frac1c}\)
\(\eqalign{ 30&=2\cdot3\cdot5\\ k^{\frac1x}&=k^{-\frac1a}\cdot k^{-\frac1b}\cdot k^{-\frac1c}\\ k^{\frac1x}&=k^{-\frac1a-\frac1b-\frac1c}\\ \dfrac1x&=-\dfrac1a-\dfrac1b-\dfrac1c\\ \dfrac1x+\dfrac1a+\dfrac1b+\dfrac1c&=\boxed{\boxed{0}} }\)
Jadi, \({\dfrac1x+\dfrac1a+\dfrac1b+\dfrac1c=0}\).
No.
Jumlah semua bilangan real x yang memenuhi persamaan- 6
- 7
- 8
- 9
- 10
ALTERNATIF PENYELESAIAN
Misal (3x − 27) = a dan (5x − 625) = b
\begin{aligned}
a^2+b^2&=(a+b)^2\\
a^2+b^2&=a^2+2ab+b^2\\
ab&=0\\
\left(3^x-27\right)\left(5^x-625\right)&=0
\end{aligned}
3 + 4 = 7
\begin{aligned}
3^x-27&=0\\
3^x&=27\\
x&=\boxed{3}
\end{aligned}
\begin{aligned}
5^x-625&=0\\
5^x&=625\\
x&=\boxed{4}
\end{aligned}
Jadi, jumlah semua bilangan real x yang memenuhi persamaan (3x − 27)2 + (5x − 625)2 = (3x + 5x − 652)2 adalah 7.
JAWAB: B
JAWAB: B
No.
Diketahui persamaan- 0
- 1
- 2
- 3
- 4
ALTERNATIF PENYELESAIAN
\begin{aligned}
2^x&=3^{x^2+x}\\
\log2^x&=\log3^{x^2+x}\\
x\log2&=\left(x^2+x\right)\log3\\
x^2\log3+x(\log3-\log2)&=0
\end{aligned}
Ada 2 akar real.
Jadi, banyak anggota penyelesaian yang mungkin adalah 2.
JAWAB: C
JAWAB: C
Post a Comment