Berikut ini adalah kumpulan soal mengenai Persamaan Eksponen tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Nilai
x yang memenuhi persamaan
{\dfrac{\sqrt[3]{(0{,}008)^{7-2x}}}{(0{,}2)^{-4x+5}}=1} adalah....
\(\eqalign{
\dfrac{\sqrt[3]{(0{,}008)^{7-2x}}}{(0{,}2)^{-4x+5}}&=1\\[4pt]
\dfrac{\sqrt[3]{\left((0{,}2)^3\right)^{7-2x}}}{(0{,}2)^{-4x+5}}&=1\\[4pt]
\dfrac{(0{,}2)^{7-2x}}{(0{,}2)^{-4x+5}}&=1\\[4pt]
(0{,}2)^{7-2x-(-4x+5)}&=1\\
(0{,}2)^{7-2x+4x-5}&=1\\
(0{,}2)^{2x+2}&=(0{,}2)^0\\
2x+2&=0\\
2x&=-2\\
x&=-1
}\)
No. 2
Hasil kali semua
x yang memenuhi persamaan
{9^{x^3-4x^2-x+4}-9^{x^2+x-6}=0} adalah....
\(\eqalign{
9^{x^3-4x^2-x+4}-9^{x^2+x-6}&=0\\
9^{x^3-4x^2-x+4}&=9^{x^2+x-6}\\
x^3-4x^2-x+4&=x^2+x-6\\
x^3-5x^2-2x+10&=0
}\)
a=1, b=-5, c=-2, d=10
\(\eqalign{
x_1x_2x_3&=-\dfrac{d}a\\
&=-\dfrac{10}1\\
&=\boxed{\boxed{-10}}
}\)
No. 3
Diberikan persamaan:
{\left(3\sqrt{\dfrac1{243}}\right)^{3x}=\left(\dfrac3{3^{x-2}}\right)^2\sqrt[3]{\dfrac19}}. Jika
x_0 memenuhi persamaan maka nilai
{1-\dfrac34x_0} adalah ....
- 1\dfrac3{16}
- 1\dfrac14
- 1\dfrac34
\(\begin{aligned}
\left(3\sqrt{\dfrac1{243}}\right)^{3x}&=\left(\dfrac3{3^{x-2}}\right)^2\sqrt[3]{\dfrac19}\\[8pt]
\left(3\sqrt{\dfrac1{3^5}}\right)^{3x}&=\left(3^{1-(x-2)}\right)^2\sqrt[3]{\dfrac1{3^2}}\\[8pt]
\left(3\sqrt{3^{-5}}\right)^{3x}&=\left(3^{1-x+2}\right)^2\sqrt[3]{3^{-2}}\\
\left(3\cdot3^{-\frac52}\right)^{3x}&=\left(3^{-x+3}\right)^2\cdot3^{-\frac23}\\
\left(3^{1-\frac52}\right)^{3x}&=3^{-2x+6}\cdot3^{-\frac23}\\
3^{-\frac92x}&=3^{-2x+\frac{16}3}\\
-\dfrac92x&=-2x+\dfrac{16}3\\
-\dfrac92x+2x&=\dfrac{16}3\\
-\dfrac52x&=\dfrac{16}3\\
x&=-\dfrac{32}{15}
\end{aligned}\)
\(\eqalign{
1-\dfrac34x_0&=1-\dfrac{\cancel{3}}{\cancel{4}}\left(-\dfrac{\cancelto{\color{red}{8}}{32}}{\cancelto{\color{red}{5}}{15}}\right)\\
&=1+\dfrac85\\
&=2\dfrac35
}\)
0 Response to "Exercise Zone : Persamaan Eksponen "
Post a Comment