Contoh Soal 4 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{l}\\ 16.& (\mathbf{\text{textrm{UM UNDIP 2012 Math IPA}}})\\ & \text{Jika}f(x)=x^3-3x^2-3x-1,\\ & \text{maka nilai dari}f(1+\sqrt[3]{2}+\sqrt[3]{4})=....\\ & \begin{array}{l}\\ \text{a}.& -\sqrt{2}\\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& \sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& =x^3-3x^2-3x-1\\ & ={(x-1)}^3-6x\\ f(1+\sqrt[3]{2}+\sqrt[3]{4})& ={(1+\sqrt[3]{2}+\sqrt[3]{4}-1)}^3-6(1+\sqrt[3]{2}+\sqrt[3]{4})\\ & ={(\sqrt[3]{2}+\sqrt[3]{4})}^3-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & ={\left(\sqrt[3]{2}(\sqrt[3]{2}+1)\right)}^3-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =2(1+3\sqrt[3]{2}+3\sqrt[3]{4}+2)-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =(6+6\sqrt[3]{2}+6\sqrt[3]{4})-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =0\end{array}\end{array}$

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

$\begin{array}{l}\\ 18.& \text{Perhatikan gambar berikut}\end{array}$

$\begin{array}{l}\\ .& \text{Persamaan grafik fungsi seperti gambar di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=2^{x-2}\\ \text{b}.& f(x)=2^x-2\\ \text{c}.& f(x)=2^x-1\\ \text{d}.& f(x)={}^2\log (x-1)\\ \text{e}.& f(x)={}^2\log (x+1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{a}& \Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\ne 0(\text{salah})\\ \text{b}& \Rightarrow (1,0)\Rightarrow f(1)=2^1-2=2^1-2=0=0(\mathbf{\text{benar}})\\ \text{c}& \Rightarrow (1,0)\Rightarrow f(1)=2^1-1=2^1-1=1\ne 0(\text{salah})\\ \text{d}& \Rightarrow (1,0)\Rightarrow f(1)={}^2\log (1-1)={}^2\log 0=\\ & \mathbf{\text{tidak mungkin}}\ne 0(\text{salah})\\ \text{e}& \Rightarrow (1,0)\Rightarrow f(1)={}^2\log (1+1)={}^2\log 2=1\ne 0(\text{salah})\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Solusi untuk persamaan}{3}^{2x+1}={81}^{x-2}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 4,5\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}{3}^{2x+1}& ={81}^{x-2}\\ \left(3^{2x}\right){.3}^1& ={\left(3^4\right)}^{x-2}\\ 3.\left(3^{2x}\right)& =3^{4x}{.3}^{-8}\\ 3.\left(3^{2x}\right)& ={\displaystyle \frac{{\left(3^{2x}\right)}^2}{3^8}}\\ 0& ={\displaystyle \frac{{\left(3^{2x}\right)}^2}{3^8}-3\left(3^{2x}\right)}\\ 0& ={\left(3^{2x}\right)}^2-3^9.\left(3^{2x}\right)\\ 0& =\left(3^{2x}\right)(\left(3^{2x}\right)-3^9)\end{array}& \begin{array}{rl}\mathbf{\text{atau}}& \\ {\left(3^{2x}\right)}^2-27.\left(3^{2x}\right)& =0\\ \left(3^{2x}\right)(\left(3^{2x}\right)-3^9)& =0\\ 3^{2x}=0\text{atau}{3}^{2x}& =3^9\\ (\text{tm})\text{atau}{3}^{2x}& =3^9\\ 2^x& =9\\ x& ={\displaystyle \frac{9}{2}}\\ \text{atau}x& =4{\displaystyle \frac{1}{2}}\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Jika}\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ & \text{maka nilai dari}{(a+b-c)}^{abc}=....\\ & \begin{array}{l}\\ \text{a}.& 1000\\ \text{b}.& 1\\ \text{c}.& 0\\ \text{d}.& -1\\ \text{e}.& -1000\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{r}\begin{array}{rl}{(\sqrt{2}+\sqrt{3}+\sqrt{5})}^2& =(\sqrt{2}+\sqrt{3}+\sqrt{5})\times (\sqrt{2}+\sqrt{3}+\sqrt{5})\\ & =\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ & +\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ & =2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ & =2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ & =10+\sqrt{2^2.6}+\sqrt{2^2.10}+\sqrt{2^2.15}\\ & =10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}& =\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ & \{\begin{array}{ll}a& =2\\ b& =3\\ c& =5\end{array}.\text{sesuai dengan urutannya}\\ \text{Sehingga nilai}& \\ {(a+b-c)}^{abc}& ={(2+3-5)}^{2.3.5}\\ & =0^{30}\\ & =0\end{array}\end{array}\end{array}$