HOTS Zone : Persamaan Eksponen

Berikut ini adalah kumpulan soal mengenai Persamaan Eksponen. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

\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

Channel Telegram

$$\begin{array}{rl}{\displaystyle \frac{8^x+{27}^x}{{12}^x+{18}^x}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{{\left(2^3\right)}^x+{\left(3^3\right)}^x}{(6\cdot 2{)}^x+(6\cdot 3{)}^x}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{2^{3x}+3^{3x}}{6^x\cdot 2^x+6^x\cdot 3^x}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{{\left(2^x\right)}^3+{\left(3^x\right)}^3}{6^x(2^x+3^x)}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{(2^x+3^x)({\left(2^x\right)}^2-\left(2^x\right)\left(3^x\right)+{\left(3^x\right)}^2)}{(2\cdot 3{)}^x(2^x+3^x)}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{{\left(2^x\right)}^2-\left(2^x\right)\left(3^x\right)+{\left(3^x\right)}^2}{\left(2^x\right)\left(3^x\right)}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{{\left(2^x\right)}^2}{\left(2^x\right)\left(3^x\right)}}-{\displaystyle \frac{\left(2^x\right)\left(3^x\right)}{\left(2^x\right)\left(3^x\right)}}+{\displaystyle \frac{{\left(3^x\right)}^2}{\left(2^x\right)\left(3^x\right)}}& ={\displaystyle \frac{7}{6}}\\ {\displaystyle \frac{2^x}{3^x}}-1+{\displaystyle \frac{3^x}{2^x}}& ={\displaystyle \frac{7}{6}}\\ {\left({\displaystyle \frac{2}{3}}\right)}^x+{\left({\displaystyle \frac{3}{2}}\right)}^x& ={\displaystyle \frac{13}{6}}\end{array}$$

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

$$\begin{array}{rl}y+{\displaystyle \frac{1}{y}}& ={\displaystyle \frac{13}{6}}\\ y^2+1& ={\displaystyle \frac{13}{6}}y\\ y^2-{\displaystyle \frac{13}{6}}y+1& =0\end{array}$$
$$\begin{array}{rl}{y}_1{y}_2& ={\displaystyle \frac{c}{a}}\\ {\left({\displaystyle \frac{2}{3}}\right)}^{x_1}{\left({\displaystyle \frac{2}{3}}\right)}^{x_2}& ={\displaystyle \frac{1}{1}}\\ {\left({\displaystyle \frac{2}{3}}\right)}^{x_1+x_2}& =1\\ x_1+x_2& =0\end{array}$$

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

No.

Diketahui sistem persamaan:
3^a+4^b=6
3^{\frac{a}b}=4
Nilai \dfrac1a+\dfrac1b= ....

Grup Telegram

$\begin{array}{rl}{3}^{\frac{a}{b}}& =4\\ {\left(3^{\frac{a}{b}}\right)}^b& =4^b\\ 3^a& =4^b\end{array}$

$\begin{array}{rl}{3}^a+4^b& =6\\ 3^a+3^a& =6\\ 2\cdot 3^a& =6\\ 3^a& =3\\ a& =1\\ {\displaystyle \frac{1}{a}}& =1\end{array}$

$\begin{array}{rl}{3}^{\frac{a}{b}}& =4\\ 3^{\frac{1}{b}}& =4\\ {\displaystyle \frac{1}{b}}& ={}^3\log 4\end{array}$

\dfrac1a+\dfrac1b=\boxed{\boxed{1+{^3\negmedspace\log4}}}

Jadi, \dfrac1a+\dfrac1b=1+{^3\negmedspace\log4}.

No.

Jika 30^x=2^{-a}=3^{-b}=5^{-c}, maka {\dfrac1x+\dfrac1a+\dfrac1b+\dfrac1c=}

Facebook

30^x=2^{-a}=3^{-b}=5^{-c}=k

30=k^{\frac1x}, 2=k^{-\frac1a}, 3=k^{-\frac1b}, 5=k^{-\frac1c}

$\begin{array}{rl}30& =2\cdot 3\cdot 5\\ k^{\frac{1}{x}}& =k^{-\frac{1}{a}}\cdot k^{-\frac{1}{b}}\cdot k^{-\frac{1}{c}}\\ k^{\frac{1}{x}}& =k^{-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}}\\ {\displaystyle \frac{1}{x}}& =-{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{c}}\\ {\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}& =\boxed{{\displaystyle \boxed{{\displaystyle 0}}}}\end{array}$

Jadi, {\dfrac1x+\dfrac1a+\dfrac1b+\dfrac1c=0}.

No.

Jumlah semua bilangan real x yang memenuhi persamaan {\left(3^x-27\right)^2+\left(5^x-625\right)^2=\left(3^x+5^x-652\right)^2} adalah ....

1. 6
2. 7
3. 8

4. 9
5. 10

C S C POSI 2021

Misal \left(3^x-27\right)=a dan \left(5^x-625\right)=b

$$\begin{array}{rl}{a}^2+b^2& =(a+b{)}^2\\ a^2+b^2& =a^2+2ab+b^2\\ ab& =0\\ (3^x-27)(5^x-625)& =0\end{array}$$

$$\begin{array}{rl}{3}^x-27& =0\\ 3^x& =27\\ x& =\boxed{{\displaystyle 3}}\end{array}$$

$$\begin{array}{rl}{5}^x-625& =0\\ 5^x& =625\\ x& =\boxed{{\displaystyle 4}}\end{array}$$

3+4=\boxed{\boxed{7}}

Jadi, jumlah semua bilangan real x yang memenuhi persamaan {\left(3^x-27\right)^2+\left(5^x-625\right)^2=\left(3^x+5^x-652\right)^2} adalah 7. JAWAB: B