HOTS Zone : Persamaan Eksponen

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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{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.

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

Grup Telegram

\(\eqalign{ 3^{\frac{a}b}&=4\\ \left(3^{\frac{a}b}\right)^b&=4^b\\ 3^a&=4^b }\)

\(\eqalign{ 3^a+4^b&=6\\ 3^a+3^a&=6\\ 2\cdot3^a&=6\\ 3^a&=3\\ a&=1\\ \dfrac1a&=1 }\)

\(\eqalign{ 3^{\frac{a}b}&=4\\ 3^{\frac1b}&=4\\ \dfrac1b&={^3\negmedspace\log4} }\)

\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}

\(\eqalign{ 30&=2\cdot3\cdot5\\ k^{\frac1x}&=k^{-\frac1a}\cdot k^{-\frac1b}\cdot k^{-\frac1c}\\ k^{\frac1x}&=k^{-\frac1a-\frac1b-\frac1c}\\ \dfrac1x&=-\dfrac1a-\dfrac1b-\dfrac1c\\ \dfrac1x+\dfrac1a+\dfrac1b+\dfrac1c&=\boxed{\boxed{0}} }\)

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

No.

Jumlah semua bilangan real x yang memenuhi persamaan {\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{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}

\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}

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