Agustus 2017

Persamaan Eksponen Bagian 3

# Persamaan Eksponen Berbentuk $A\left \{ a^{f\left ( x \right )} \right \}^{2}+B\left \{a^{f\left ( x \right )} \right \}+C=0$

Cara untuk menyelesaikan persamaan eksponen berbentuk $A\left \{ a^{f\left ( x \right )} \right \}^{2}+B\left \{a^{f\left ( x \right )} \right \}+C=0$ :
-- Misalkan $a^{f\left ( x \right )}=y$ , maka persamaan semula menjadi $Ay^{2}+By+c=0$ .
-- Selesaikan persamaan kuadrat dalam

, didapat dua akar real, ambil akar real positif.
-- Akar-akar tersebut disubstitusikan ke persamaan $a^{f\left ( x \right )}=y$ .

Soal dan Pembahasan

1. SPMB IPA Regional II, 2003

$\begin {array}{lcl} \textup{Jika}\: 3^{x+2} +9^{x+1}=810,\: \textup{maka}\: 3^{x-3}\: \: \textup{sama dengan}\: .....\\ \textup{A}.\: \frac{1}{9}\\ \textup{B}.\: \frac{1}{3}\\ \textup{C}.\: 1\\ \textup{D}.\: 3\\ \textup{E}.\: 9 \end{array}$

Pembahasan:

$\begin {array}{lcl} 3^{x+2} +9^{x+1}&=&810\\ 3^{x}\cdot 3^{2}+9^{x}\cdot 9&=&810\\ 9\cdot 3^{x}+\left ( 3^{2} \right )^{x}\cdot 9&=&810\\ 9\cdot 3^{x}+\left ( 3^{x} \right )^{2}\cdot 9&=&810\\ \textup{misal}\: 3^{x}=y,\: \textup{maka}\\ 9y+9y^{2}&=&810\\ 9y^{2}+9y-810&=&0\: \: \: \: \: \: \: \: \: \: \: \: \left ( \textup{sama-sama dibagi 9} \right )\\ y^{2}+y-90&=&0\\ \left ( y+10 \right )\left ( y-9 \right )&=&0\\ y=-10\: \: \textup{atau}\: \:y=9 \end{array}$
$\begin {array}{lcl} \textup{ambil yang positif},\: y=9\\ y=3^{x}\\ 9=3^{x}\\ \textup{maka}\\ 3^{x-3}=3^{x}\cdot 3^{-3}\\ \: \: \: \: \: \: \: \: \: \, =\frac{3^{x}}{3^{3}}\\ \: \: \: \: \: \: \: \: \: \, =\frac{9}{27}\\ \: \: \: \: \: \: \: \: \: \, =\frac{1}{3} \end{array}$

Jawaban _______________________________________ (B)

2. SPMB IPA Regional III, 2003

   $\begin {array}{lcl} \textup{Jumlah akar-akar persamaan}\: \: 5^{x+1}+5^{2-x}-30=0\: \textup{adalah}\: .....\\ \textup{A}.\: -2\\ \textup{B}.\: -1\\ \textup{C}.\: \: \: 0\\ \textup{D}.\:\: \: 1\\ \textup{E}.\:\: \: 2 \end{array}$

Pembahasan:

$\begin {array}{lcl} 5^{x+1}+5^{2-x}-30&=&0\\ 5^{x}\cdot 5+5^{2}\cdot 5^{-x}-30&=&0\\ 5^{x}\cdot 5+\frac{5^{2}}{5^{x}}-30&=&0\: \: \: \: \: \: \left ( \textup{sama-sama dibagi}\: \: 5^{x} \right )\\ 5^{x}\cdot 5^{x}\cdot 5+5^{2}-30\cdot 5^{x}&=&0\\ 5^{2x}\cdot 5+25-30\cdot 5^{x}&=&0\\ 5\cdot \left (5^{x} \right )^{2}-30\cdot \left ( 5^{x} \right )+25&=&0\\ \textup{misal}\: 5^{x}=y,\: \textup{maka}\\ 5y^{2}-30y+25&=&0\; \; \; \; \; \left ( \textup{sama-sama dibagi 5} \right )\\ y^{2}-6y+5&=&0\\ \left ( y-1 \right )\left ( y-5 \right )&=&0 \end{array}$
$\begin {array}{lcl} y\; \: =1\; \textup{atau}\; \: \: y\: =5\\ 5^{x}=1\: \textup{atau}\; \: \: 5^{x}=5\\ 5^{x}=5^{0}\: \textup{atau}\: \: 5^{x}=5^{1}\\ x\; \, =0\; \textup{atau}\: \; \; \: x=1\\ \textup{Jumlah akar-akar persamaannya = 0+1}\\ \textup{ \: \: \: \: \: \: \: \: \: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; = 1 } \end{array}$

Jawaban _______________________________________ (D)

CARA PRAKTIS

$\begin {array}{lcl} A^{px+q}+A^{r-px} &=&C\\\\ x_{1}+x_{2}&=&\frac{r-q}{p} \end{array}$

$\begin {array}{lcl} 5^{x+1}+5^{2-x}-30 &=&0\\ 5^{x+1}+5^{2-x} &=&30\\ x_{1}+x_{2}&=&\frac{2-1}{1}\\ &=&1 \end{array}$

Jawaban _______________________________________ (D)

3. UN 2007

$\begin{array}{lcl} \textup{Akar-akar persamaan}\; 3^{2x+1}-28\cdot 3^{x}+9=0\; \textup{adalah}\; x_{1}\; \textup{dan}\; x_{2},\\ \textup{di mana}\: x_{1}> x_{2},\; \textup{maka}\; 3x_{1}-x_{2}=.....\\ \textup{A}.\; -5\\ \textup{B}.\; -1\\ \textup{C}.\; \; 4\\ \textup{D}.\; \; 5\\ \textup{E}.\; \; 6 \end{array}$

Pembahasan:

$\begin{array}{lcl} 3^{2x+1}-28\cdot 3^{x}+9&=&0\\ 3^{2x}\cdot 3-28\cdot 3^{x}+9&=&0\\ 3\left ( 3^{x} \right )^{2}-28\left ( 3^{x} \right )+9&=&0\\\\ \textup{misal}\; y=3^{x},\textup{maka}\\\\ 3y^{2}-28y+9&=&0\\ \left ( 3y-1 \right )\left ( y-9 \right )&=&0\\ \end{array}$

$\begin{array}{lcl} 3y=1 \; \textup{atau}\; y=9\\ y=\frac{1}{3}\; \; \textup{atau}\; y=9\\ 3^{x}=\frac{1}{3}\; \textup{atau}\; 3^{x}=9\\ 3^{x}=3^{-1}\; \textup{atau}\; 3^{x}=3^{2}\\ x=-1\; \textup{atau}\; x=2\\ \textup{karena}\; x_{1}> x_{2}, \textup{maka}\\ x_{1}=2\\ x_{2}=-1\\\\ 3x_{1}-x_{2}=3\left ( 2 \right )-1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; =5 \end{array}$

Jawaban _______________________________________ (D)

4. SIMAK UI 2009 Kode 912

   $\begin{array}{lcl} \textup{Jika}\; x_{1}\; \textup{dan}\; x_{2}\; \textup{merupakan akar-akar persamaan }\\ 5^{x+1}+5^{2-x}=126,\; \textup{maka}\; x_{1}+x_{2}=\: .....\\ \textup{A}.\; \; 25\frac{1}{5}\\ \textup{B}.\; \; 5\\ \textup{C}.\; \; 1\\ \textup{D}.-1\\ \textup{E}.-3 \end{array}$

Pembahasan:

$\begin{array}{lcl} 5^{x+1}+5^{2-x}&=&126\\ 5^{x}\cdot 5+5^{2}\cdot 5^{-x}-126&=&0\\ 5^{x}\cdot 5+\frac{5^{2}}{5^{x}}-126&=&0\; \; \; \; \; \; \; \; \; \left ( \textup{sama-sama dikali}\; 5^{x} \right )\\ 5^{x}\cdot 5^{x}\cdot 5+5^{^{2}}-126\cdot 5^{x}&=&0\\ 5\left (5^{2x} \right )-126\left ( 5^{x} \right )+25&=&0\\ 5\left ( 5^{x} \right )^{2}-126\left ( 5^{x} \right )+25&=&0\\\\\textup{misal } \: y=5^{x},\; \textup{maka}\\ 5y^{2}-126y+25&=&0\\ \left ( 5y-1 \right )\left ( y-25 \right )&=&0\\ \end{array}$
$\begin{array}{lcl} 5y=1\; \textup{atau}\; y=25\\ y=\frac{1}{5}\; \textup{atau}\; y=25\\ 5^{x}=\frac{1}{5}\; \textup{atau}\; 5^{x}=25\\ 5^{x}=5^{-1}\; \textup{atau}5^{x}=5^{2}\\ x=-1\; \textup{atau}\; x=2\\ \textup{karena tidak ada syarat untuk}\; x_{1}\; \textup{dan}\; x_{2},\\ \textup{maka bebas menentukan nilai}\; x_{1}\; \textup{dan}\; x_{2}\\ x_{1}=-1\\ x_{2}=2\\ x_{1}+x_{2}=-1+2\\ \; \; \; \; \; \; \; \; \; \; \; \; =1 \end{array}$

Jawaban _______________________________________ (C)

CARA PRAKTIS

$\begin{array}{lcl} A^{px+q}+A^{r-px}=C\\ \textup{koef \textit{p} harus sama}\\ \\x_{1}+x_{2}=\frac{r-q}{p} \end{array}$

$\begin{array}{lcl} 5^{x+1}+5^{2-x}=126\\ \textup{koef \textit{p} nilainya sama yaitu 1, maka rumus praktis tersebut bisa diterapkan}\\\\ x_{1}+x_{2}=\frac{2-1}{1}\\ \; \; \; \; \; \; \; \; \; \; \; \; =1 \end{array}$

Jawaban _______________________________________ (C)

5. SBMPTN 2014 Kode 622 dan 673
   $\begin{array} {lcl} \textup{Jika}\: 4^{x}-4^{x-1}=6,\; \textup{maka}\; \left ( 2x \right )^{x}\; \textup{sama dengan .....}\\ \textup{A}.\; 3\\ \textup{B}.\; 3\sqrt{3}\\ \textup{C}.\; 9\\ \textup{D}.\; 9\sqrt{3}\\ \textup{E}.\; 27 \end{array}$

Pembahasan:

$\begin{array} {lcl} 4^{x}-4^{x-1}&=&6\\ 4^{x}-4^{x}\cdot 4^{-1}&=&6\\ 4^{x}-\frac{4^{x}}{4}&=&6\; \; \; \; \; \; \left ( \textup{sama-sama di kali 4} \right )\\ 4\cdot 4^{x}-4^{x}&=&6\cdot 4\\ 3\cdot 4^{x}&=&24\\ 4^{x}&=&8\\ 2^{2x}&=&2^{3}\\ 2x&=&3\\ x&=&\frac{3}{2}\\\\ \left ( 2x \right )^{x}&=&\left ( 3 \right )^{\frac{3}{2}}\\ &=&3\cdot 3^{\frac{1}{2}}\\ &=&3\sqrt{3} \end{array}$

Jawaban _______________________________________ (B)

6. UN 2009 P3
$\begin{array} {lcl} \textup{Akar-akar persamaan}\; 2^{x}+2^{3-x}=9\; \textup{adalah }\; \alpha \; \textup{dan} \; \beta .\; \textup{Nilai}\; \alpha +\beta =\; .....\\ \textup{A}.\; 3\\ \textup{B}.\; 4\\ \textup{C}.\; 6\\ \textup{D}.\; 8\\ \textup{E}.\; 9 \end{array}$

Pembahasan:

$\begin{array} {lcl} 2^{x}+2^{3-x}&=&9\\ 2^{x}+2^{3}\cdot 2^{-x}&=&9\\ 2^{x}+\frac{8}{2^{x}}&=&9\; \; \; \; \; \; \; \; \left ( \textup{sama-sama di kali}\; 2^{x} \right )\\ 2^{x}\cdot 2^{x}+8&=&9\cdot 2^{x}\\ 2^{2x}-9\cdot 2^{x}+8&=&0\\ \left ( 2^{x} \right )^{2}-9\left ( 2^{x} \right )+8&=&0\\\\ \textup{Misal}\; y=2^{x},\; \textup{maka}\\ y^{2}-9y+8&=&0\\ \left ( y-1 \right )\left ( y-8 \right )&=&0 \end{array}$
$\begin{array} {lcl} y=1\; \textup{atau}\; y=8\\ 2^{x}=1\; \textup{atau}\; 2^{x}=8\\ 2^{x}=2^{0}\; \textup{atau}\; 2^{x}=2^{3}\\ x=0\; \textup{atau}\; x=3,\; \textup{maka}\\ \alpha =0\\ \beta =3\\ \alpha +\beta=0+3\\ \; \; \; \; \; \; \; \; \; \; =3 \end{array}$

Jawaban _______________________________________ (A)

CARA PRAKTIS

$\begin{array} {lcl} A^{px+q}+A^{r-px}=C\\ \textup{koef }\; p\; \textup{harus sama}\\\\ x_{1}+x_{2}=\frac{r-q}{p}\\\\ 2^{x}+2^{3-x}&=&9\\ 2^{x+0}+2^{3-x}&=&9\\ \alpha +\beta =\frac{3-0}{1}\\ \; \; \; \; \; \; \; \; \; \; =3 \end{array}$

Jawaban _______________________________________ (A)

7. SIMAK UI 2013 Kode 333
   $\begin{array}{lcl} \textup{Jika diketahui bahwa}\; 2^{cos2x}+2^{cos^{2}x}=3\cdot 2^{-cos2\pi }\; \textup{nilai \textit{x} adalah .....}\\ \textup{1}.\; \frac{\pi }{2}\\ 2.\; \frac{\pi }{3}\\ 3.\; \frac{3\pi }{2}\\ 4.\; \pi \end{array}$

Pembahasan:

$\begin{array}{lcl} 2^{cos2x}+2^{cos^{2}x}&=&3\cdot 2^{-cos2\pi }\; \; \; \; \; cos2\pi =cos2\left ( 180^{0} \right )=cos360^{0}=1\\ 2^{2cos^{^{2}}x-1}+2^{cos^{2}x}&=&3\cdot 2^{-1}\\ 2^{2cos^{2}x}\cdot 2^{-1}+2^{cos^{2}x}&=&\frac{3}{2}\\ \left ( 2^{cos^{2}x} \right )^{2}\cdot \frac{1}{2}+\left ( 2^{cos^{2}x} \right )&=&\frac{3}{2}\; \; \; \; \; \; \; \; \left (\textup{sama-sama di kali}\; 2 \right )\\ \left ( 2^{cos^{2}x} \right )^{2}+2\left ( 2^{cos^{2}x} \right )&=&\frac{3}{2}\cdot 2\\ \left ( 2^{cos^{2}x} \right )^{2}+2\left ( 2^{cos^{2}x} \right )&=&3\\ \left ( 2^{cos^{2}x} \right )^{2}+2\left ( 2^{cos^{2}x} \right )-3&=&0 \end{array}$
$\begin{array}{lcl} \textup{Misal}\; y=2^{cos^{2}x}\;\textup{maka}\\ y^{2}+2y-3&=&0\\ \left ( y+3 \right )\left ( y-1 \right )&=&0\\ y=-3\; \textup{atau}\; y=1\\ \textup{syarat akar positif}\\ \textup{yang memenuhi nilai \textit{y}=1}\\ 2^{cos^{2}x}=1\\ 2^{cos^{2}x}=2^{0}\\ cos^{2}x=0\\ \; \: cosx=0 \end{array}$
$\begin{array}{lcl} \; \: cosx=0\; \; \; \rightarrow \; cosx=cos\frac{\pi }{2}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x_{1}=\frac{\pi }{2}\\\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; cosx=cos\frac{3\pi }{2}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x_{2}=\frac{3\pi }{2} \end{array}$

Jawaban yang benar 2 dan 3

Jawaban _______________________________________ (B)

8. UN 2007 P14

   $\begin{array}{lcl} \textup{Diketahui }\; x_{1}\; \textup{dan}\; x_{2}\; \textup{akar-akar persamaan}\\ 9^{x}-\frac{10}{3}\cdot 3^{x}+1=0.\\ \textup{Nilai}\; x_{1}+x_{2}=\; .....\\ \textup{A}.\; 2\\ \textup{B}.\; \frac{3}{2}\\ \textup{C}.\; 1\\ \textup{D}.\; 0\\ \textup{E}.\; -2 \end{array}$

Pembahasan:

$\begin{array}{lcl} 9^{x}-\frac{10}{3}\cdot 3^{x}+1&=&0\\ \left ( 3^{2} \right )^{x}-\frac{10}{3}\left ( 3^{x} \right )+1&=&0\; \; \; \; \; \left ( \textup{sama-sama di kali 3} \right )\\ 3\left ( 3^{x} \right )^{2}-10\left ( 3^{x} \right )+3&=&0\\\\ \textup{Misal}\; y=3^{x},\; \textup{maka}\\ 3y^{2}-10y+3&=&0\\ \left ( 3y-1 \right )\left ( y-3 \right )&=&0 \end{array}$
$\begin{array}{lcl} \left ( 3y-1 \right )\left ( y-3 \right )&=&0\\ y=\frac{1}{3}\; \textup{atau}\; y=3\\ 3^{x}=\frac{1}{3}\; \textup{atau}\; 3^{x}=3\\ 3^{x}=3^{-1}\; \textup{atau}\; 3^{x}=3\\ x=-1\; \textup{atau}\; x=1\\ \textup{karena tidak ada syarat untuk nilai}\; x_{1}\; \textup{dan}\; x_{2}\\ \textup{maka bebas untuk menentukan nilainya}\\ x_{1}=-1\\ x_{2}=1\\ x_{1}+x_{2}=-1+1\\ \; \; \; \; \; \; \; \; \; \; \; \; =0 \end{array}$

Jawaban _______________________________________ (D)

9. UN 2008 P03

   $\begin{array}{lcl} \textup{Akar-akar persamaan}\; 4^{x}-12\cdot 2^{x}+32=0\; \textup{adalah}\; x_{1}\; \textup{dan}\: x_{2}.\\ \textup{Nilai}\; x_{1}+x_{2}=\; .....\\ \textup{A}.\; 3\\ \textup{B}.\; 6\\ \textup{C}.\;8\\ \textup{D}.\; 12\\ \textup{E}.\; 32\\ \end{array}$

Pembahasan:

$\begin{array}{lcl} 4^{x}-12\cdot 2^{x}+32&=&0\\ \left ( 2^{2} \right )^{x}-12\left ( 2^{x} \right )+32&=&0\\ \left ( 2^{x} \right )^{2}-12\left ( 2^{x} \right )+32&=&0\\\\ \textup{Misal}\; y=2^{x},\; \textup{maka}\\ y^{2}-12y+32&=&0\\ \left ( y-4 \right )\left ( y-8 \right )&=&0 \end{array}$
$\begin{array}{lcl} y=4\; \textup{atau}\; y=8\\ 2^{x}=4\; \textup{atau}\; 2^{x}=8\\ 2^{x}=2^{2}\; \textup{atau}\; 2^{x}=2^{3}\\ x=2\; \textup{atau}\; x=3\\\\ \textup{karena tidak ada syarat untuk nilai}\; x_{1}\; \textup{dan}\; x_{2}\\ \textup{maka kita bebas menentukan nilainya}\\ x_{1}=2\\ x_{2}=3\\ x_{1}\cdot x_{2}=2\cdot 3\\ \; \; \; \; \; \; \; \; \; \; \: =6 \end{array}$

Jawaban _______________________________________ (B)

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