Contoh Soal 4 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{ll}\\ 16.&(\textbf{\textrm{UM UNDIP 2012 Math IPA}})\\ &\textrm{Jika}\: \: f(x)=x^{3}-3x^{2}-3x-1,\\ &\textrm{maka nilai dari}\: \: f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-\sqrt{2}\\ \textrm{b}.&-1\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&1\\ \textrm{e}.&\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=x^{3}-3x^{2}-3x-1\\ &=\left ( x- 1\right )^{3}-6x\\ f\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )&=\left (1+\sqrt[3]{2}+\sqrt[3]{4}-1 \right )^{3}-6\left ( 1+\sqrt[3]{2}+\sqrt[3]{4} \right )\\ &=\left ( \sqrt[3]{2}+\sqrt[3]{4} \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left (\sqrt[3]{2} \left ( \sqrt[3]{2}+1 \right ) \right )^{3}-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=2\left ( 1+3\sqrt[3]{2}+3\sqrt[3]{4}+2 \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=\left ( 6+6\sqrt[3]{2}+6\sqrt[3]{4} \right )-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{l}\\ 17.&\textrm{Jika}\: \: \textbf{a}\: \: \textrm{dan}\: \: \textbf{b}\: \: \textrm{adalah bilangan bulat positif}\\ &\textrm{yang memenuhi persamaan}\: \: \textbf{a}^{\textbf{b}}=2^{20}-2^{19},\\ & \textrm{maka nilai dari}\: \: \textbf{a}+\textbf{b}=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&3\\ \textrm{b}.&7\\ \textrm{c}.&19\\ \color{red}\textrm{d}.&21\\ \textrm{e}.&23 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textbf{a}^{\textbf{b}}&=2^{20}-2^{19}\\ &=2^{^{^{(19+1)}}}-2^{19}\\ &=2^{19}.2^{1}-2^{19}\\ &=2^{19}(2-1)\\ &=2^{19}\\ \textrm{Se}&\textrm{hingga nilai}\: \: \textbf{a}+\textbf{b}=2+19=21\end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Perhatikan gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\quad\: \, &\textrm{Persamaan grafik fungsi seperti gambar di atas adalah}\, ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f(x)=2^{x-2}\\ \color{red}\textrm{b}.&f(x)=2^{x}-2\\ \textrm{c}.&f(x)=2^{x}-1\\ \textrm{d}.&f(x)=\, ^{2}\log (x-1)\\ \textrm{e}.&f(x)=\, ^{2}\log (x+1) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{a}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\neq 0\: \: (\textrm{salah})\\ \textrm{b}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-2=2^{1}-2=0= 0\: \: (\textbf{benar})\\ \textrm{c}&\Rightarrow (1,0)\Rightarrow f(1)=2^{1}-1=2^{1}-1=1\neq 0\: \: (\textrm{salah})\\ \textrm{d}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1-1)=\, ^{2}\log 0=\\ &\qquad\qquad\color{magenta}\textbf{tidak mungkin}\neq 0\: \: (\textrm{salah})\\ \textrm{e}&\Rightarrow (1,0)\Rightarrow f(1)=\, ^{2}\log (1+1)=\, ^{2}\log 2=1\neq 0\: \: (\textrm{salah})\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Solusi untuk persamaan}\: \: 3^{2x+1}=81^{x-2}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&0\\ \textrm{b}.&2\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&4,5\\ \textrm{e}.&16 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{array}{|c|c|}\hline \begin{aligned}3^{2x+1}&=81^{x-2}\\ \left (3^{2x} \right ).3^{1}&=\left (3^{4} \right )^{x-2}\\ 3.\left (3^{2x} \right )&=3^{4x}.3^{-8}\\ 3.\left (3^{2x} \right )&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}\\ 0&=\displaystyle \frac{\left (3^{2x} \right )^{2}}{3^{8}}-3\left ( 3^{2x} \right )\\ 0&=\left ( 3^{2x} \right )^{2}-3^{9}.\left ( 3^{2x} \right )\\ 0&=\left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right ) \end{aligned} &\begin{aligned}\textbf{atau}\qquad\qquad\qquad&\\ \left ( 3^{2x} \right )^{2}-27.\left ( 3^{2x} \right )&=0\\ \left (3^{2x} \right )\left ( \left ( 3^{2x} \right )-3^{9} \right )&=0\\ 3^{2x}=0\: \: \textrm{atau}\: \: 3^{2x}&=3^{9}\\ (\textrm{tm})\quad \textrm{atau}\: \: 3^{2x}&=3^{9}\\ 2^{x}&=9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{atau}\: \: x&=4\displaystyle \frac{1}{2} \end{aligned} \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ &\textrm{maka nilai dari}\: \: \left ( a+b-c \right )^{abc}=....\\ &\begin{array}{lllllllll}\\ \textrm{a}.&1000\\ \textrm{b}.&1\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&-1\\ \textrm{e}.&-1000 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\begin{aligned}\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )^{2}&=\left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\times \left (\sqrt{2}+\sqrt{3}+\sqrt{5} \right )\\ &=\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ &\qquad\qquad+\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ &=2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ &=2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ &=10+\sqrt{2^{2}.6}+\sqrt{2^{2}.10}+\sqrt{2^{2}.15}\\ &=10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}&=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ &\begin{cases} a &=2 \\ b & =3 \\ c &=5 \end{cases}.\qquad \textrm{sesuai dengan urutannya}\\ \textrm{Sehingga nilai}\qquad &\\ \left ( a+b-c \right )^{abc}&=\left ( 2+3-5 \right )^{2.3.5}\\ &=0^{30}\\ &=0 \end{aligned} \end{aligned} \end{array}$