Contoh Soal 9 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 41.& \text{Penyelesaian pertidaksamaan eksponen}\\ & {\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}>\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x>{\displaystyle \frac{6}{5}}\\ \text{b}.& x<-{\displaystyle \frac{6}{5}}\\ \text{c}.& x>{\displaystyle \frac{5}{6}}\\ \text{d}.& x<-2\\ \text{e}.& x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}& >\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\\ 3^{-(2x+1)}& >3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)& >{\displaystyle \frac{1}{2}(3-(x-1))}\\ -4x-2& >4-x\\ -4x+x& >4+2\\ -3x& >6\\ 3x& <-6\\ x& <-2\end{array}\end{array}$

$\begin{array}{l}\\ 42.& (\mathbf{\text{UMPTN 01}})\text{Nilai}x\text{yang memenuhi}\\ & 4^{x^2-x-2}{.2}^{x^2+3x-10}<{\displaystyle \frac{1}{16}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-5\text{atau}x>-2\\ \text{b}.& x<-2\text{atau}x>{\displaystyle \frac{5}{3}}\\ \text{c}.& -2<x<-1\\ \text{d}.& -2<x<{\displaystyle \frac{5}{3}}\\ \text{e}.& -5<x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{4}^{x^2-x-2}{.2}^{x^2+3x-10}& <{\displaystyle \frac{1}{16}}\\ 2^{2(x^2-x-2)+(x^2+3x-10)}& <2^{-4}\\ 2(x^2-x-2)+(x^2+3x-10)& <-4\\ 3x^2-2x+3x-4-10+4& <0\\ 3x^2+x-10+& <0\\ (x+2)(3x-5)& <0\\ \therefore -2<x& <{\displaystyle \frac{5}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 43.& (\mathbf{\text{SPMB 04 Mat IPA}})\text{Himpunan Penyelesaian}\\ & \text{pertidaksamaan eksponen}\\ & 2\sqrt{4^{x^2-3x+2}}<\sqrt[3]{{\left({\displaystyle \frac{1}{2}}\right)}^{3-6x}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x>4\}\\ \text{b}.& \{x|x>2\}\\ \text{c}.& \{x|x<1\}\\ \text{d}.& \{x|1<x<4\}\\ \text{e}.& \{x|2\le x\le 3\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}2\sqrt{4^{x^2-3x+2}}& <\sqrt[3]{{\left({\displaystyle \frac{1}{2}}\right)}^{3-6x}}\\ 2^{1+\frac{2}{2}(x^2-3x+2)}& <2^{-\frac{3-6x}{3}}\\ 1+(x^2-3x+2)& <-{\displaystyle \frac{3-6x}{3}}\\ x^2-3x+3& <-1+2x\\ x^2-5x+4& <0\\ (x-1)(x-4)& <0\\ 1<x& <4\end{array}\end{array}$

$\begin{array}{l}\\ 44.& (\mathbf{\text{SBMPTN 2015 Mat IPA}})\\ & \text{Nilai}c\text{yang memenuhi}\\ & (0,12{)}^{4x^2+8x+c}<(0,0144{)}^{x^2+4x+4}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& c>0\\ \text{b}.& c>2\\ \text{c}.& c>4\\ \text{d}.& c>6\\ \text{e}.& c>8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}(0,12{)}^{4x^2+8x+c}& <(0,0144{)}^{x^2+4x+4}\\ (0,12{)}^{4x^2+8x+c}& <(0,12{)}^{2(x^2+4x+4)}\\ 4x^2+8x+c& >2(x^2+4x+4)\\ 2x^2+c-8& >0\text{haruslah definit positif}\\ \text{Syaratnya}& \{\begin{array}{ll}a& =2>0\\ D& =b^2-4ac<0\end{array}\\ \text{Maka}D& =b^2-4ac<0\\ \mathbf{\text{ambil dari}}& 2x^2-c-8=0\{\begin{array}{ll}a& =2\\ b& =0\\ c& =c-8\end{array}\\ & =0^2-4.2(c-8)<0\\ -8c& +64<0\\ -8c& <-64\\ 8c& >64\\ c& >8\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Nilai}x\text{yang memenuhi}\\ & 9^{2x}-{10.9}^x+9>0,x\in \mathbb{R}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<1\text{atau}x>0\\ \text{b}.& x<0\text{atau}x>1\\ \text{c}.& x<-1\text{atau}x>2\\ \text{d}.& x<1\text{atau}x>2\\ \text{e}.& x<-1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{9}^{2x}-{10.9}^x+9& >0\\ {\left(9^x\right)}^2-10.\left(9^x\right)+9& >0\\ (9^x-1)(9^x-9)& >0\\ 9^x<1\text{atau}{9}^x& >9\\ 9^x<9^0\text{atau}{9}^x& >9^1\\ x<0\text{atau}x& >1\end{array}\end{array}$