Contoh Soal 9 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{ll}\\ 41.&\textrm{Penyelesaian pertidaksamaan eksponen}\\ &\left ( \displaystyle \frac{1}{3} \right )^{2x+1}>\sqrt{\displaystyle \frac{27}{3^{x-1}}} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x>\displaystyle \frac{6}{5}\\ \textrm{b}.&x<-\displaystyle \frac{6}{5}\\ \textrm{c}.&x>\displaystyle \frac{5}{6}\\ \color{red}\textrm{d}.&x<-2\\ \textrm{e}.&x<2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{1}{3} \right )^{2x+1}&>\sqrt{\displaystyle \frac{27}{3^{x-1}}}\\ 3^{-(2x+1)}&>3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)&>\displaystyle \frac{1}{2}(3-(x-1))\\ -4x-2&>4-x\\ -4x+x&>4+2\\ -3x&>6\\ 3x&<-6\\ x&<-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&(\textbf{UMPTN 01})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &4^{x^{2}-x-2}.2^{x^{2}+3x-10}<\displaystyle \frac{1}{16} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-5\: \: \textrm{atau}\: \: x>-2\\ \textrm{b}.&x<-2\: \: \textrm{atau}\: \: x>\displaystyle \frac{5}{3}\\ \textrm{c}.&-2<x<-1\\ \color{red}\textrm{d}.&-2<x<\displaystyle \frac{5}{3}\\ \textrm{e}.&-5<x<2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}4^{x^{2}-x-2}.2^{x^{2}+3x-10}&<\displaystyle \frac{1}{16}\\ 2^{2\left ( x^{2}-x-2 \right )+\left ( x^{2}+3x-10 \right )}&<2^{-4}\\ 2\left ( x^{2}-x-2 \right )+\left ( x^{2}+3x-10 \right )&<-4\\ 3x^{2}-2x+3x-4-10+4&<0\\ 3x^{2}+x-10+&<0\\ (x+2)(3x-5)&<0\\ \therefore \qquad-2<x&<\displaystyle \frac{5}{3} \end{aligned} \end{array}$

$\begin{array}{l}\\ 43.&(\textbf{SPMB 04 Mat IPA})\textrm{Himpunan Penyelesaian}\\ & \textrm{pertidaksamaan eksponen}\\ &2\sqrt{4^{x^{2}-3x+2}}<\sqrt[3]{\left (\displaystyle \frac{1}{2} \right )^{3-6x}} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x> 4 \right \}\\ \textrm{b}.&\left \{ x|x> 2 \right \}\\ \textrm{c}.&\left \{ x|x<1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|1<x<4 \right \}\\ \textrm{e}.&\left \{ x|2\leq x\leq 3 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}2\sqrt{4^{x^{2}-3x+2}}&<\sqrt[3]{\left (\displaystyle \frac{1}{2} \right )^{3-6x}}\\ 2^{1+\frac{2}{2}\left ( x^{2}-3x+2 \right )}&<2^{- \frac{3-6x}{3}}\\ 1+\left ( x^{2}-3x+2 \right )&<-\displaystyle \frac{3-6x}{3}\\ x^{2}-3x+3&<-1+2x\\ x^{2}-5x+4&<0\\ (x-1)(x-4)&<0\\ 1<x&<4 \end{aligned} \end{array}$

$\begin{array}{l}\\ 44.&(\textbf{SBMPTN 2015 Mat IPA})\\ &\textrm{Nilai}\: \: c\: \: \textrm{yang memenuhi}\\ &(0,12)^{4x^{2}+8x+c}<(0,0144)^{x^{2}+4x+4} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&c>0\\ \textrm{b}.&c>2\\ \textrm{c}.&c>4\\ \textrm{d}.&c>6\\ \color{red}\textrm{e}.&c>8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}(0,12)^{4x^{2}+8x+c}&<(0,0144)^{x^{2}+4x+4}\\ (0,12)^{4x^{2}+8x+c}&<(0,12)^{2\left (x^{2}+4x+4 \right )}\\ 4x^{2}+8x+c&>2\left (x^{2}+4x+4 \right )\\ 2x^{2}+c-8&>0\quad \color{magenta}\textrm{haruslah definit positif}\\ \textrm{Syaratnya}&\begin{cases} a &=2>0 \\ D &=b^{2}-4ac<0\\ \end{cases}\\ \textrm{Maka}\quad D&=b^{2}-4ac<0\\ \textbf{ambil dari}&\: \: \: 2x^{2}-c-8=0\begin{cases} a &=2 \\ b &=0 \\ c &=c-8 \end{cases}\\ &=0^{2}-4.2(c-8)<0\\ -8c&+64<0\\ -8c&<-64\\ 8c&>64\\ c&>8 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &9^{2x}-10.9^{x}+9>0, \: \: x\in \mathbb{R}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x<1\: \: \textrm{atau}\: \: x>0\\ \color{red}\textrm{b}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{e}.&x<-1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}9^{2x}-10.9^{x}+9&>0\\ \left (9^{x} \right )^{2}-10.\left ( 9^{x} \right )+9&>0\\ \left ( 9^{x}-1 \right )\left ( 9^{x}-9 \right )&>0\\ 9^{x}<1\: \: \textrm{atau}\: \: 9^{x}&>9\\ 9^{x}<9^{0}\: \: \textrm{atau}\: \: 9^{x}&>9^{1}\\ x<0\: \: \textrm{atau}\: \: x&>1 \end{aligned} \end{array}$