Latihan Soal 5 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

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

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

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

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

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

$\begin{array}{l}\\ 46.& \text{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -2& & & \\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{5}^{x+1}+5^{2-x}-30& =0\\ \left(5^x\right){.5}^1+{\displaystyle \frac{5^2}{5^x}-30}& =0\\ 5{\left(5^x\right)}^2+25-30\left(5^x\right)& =0\\ \text{Persamaan kuadrat}& \text{dalam}{5}^x,\text{maka}\\ 5(5^x{)}^2-30(5^x)+25& =0\{\begin{array}{ll}a& =5\\ b& =-30\\ c& =25\end{array}\\ (5^{x_1}).\left(5^{x_2}\right)& ={\displaystyle \frac{c}{a}}\\ 5^{x_1+x_2}& ={\displaystyle \frac{25}{5}=5}\\ 5^{x_1+x_2}& =5^1\\ x_1+x_2& =1\end{array}\end{array}$

$\begin{array}{l}\\ 47.& \text{Jumlah akar-akar persamaan}\\ & {2020}^{x^2-7x+7}={2021}^{x^2-7x+7}\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -7\\ \text{b}.& -5\\ \text{c}.& -3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{2020}^{x^2-7x+7}& ={2021}^{x^2-7x+7}\\ \text{Karena basis}& \text{tidak sama},\\ \text{maka harusl}& \text{ah pangkatnya}=0,\\ x^2-7x+7& =0\\ \text{dan jumlah}& \text{akar-akarnya adalah}:\\ x_1+x_2& =-{\displaystyle \frac{b}{a},\text{dari persamaan}}\\ x^2-7x+7& =0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =7\end{array}\\ \text{maka}{x}_1+x_2& =-{\displaystyle \frac{b}{a}=-\frac{-7}{1}=7}\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Nilai dari}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 5\\ \text{c}.& 10\\ \text{d}.& 20\\ \text{e}.& 40\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}}& ={\displaystyle \frac{2^4{.2}^{2016}+2^2{.2}^{2016}}{2^2{.2}^{2018}+2^{2016}}}\\ & ={\displaystyle \frac{2^{2016}(2^4+2^2)}{2^{2016}(2^2+1)}}\\ & ={\displaystyle \frac{16+4}{4+1}}\\ & ={\displaystyle \frac{20}{5}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 49.& (\mathbf{\text{UM IPB}})\text{Jika}ab=a^b\text{dan}{\displaystyle \frac{a}{b}=a^{3b}}\\ & \text{maka nilai}a\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 0,5\\ \text{c}.& 1\\ \text{d}.& 0,25\\ \text{e}.& 0,75\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \\ ab& =a^b\\ b& ={\displaystyle \frac{a^b}{a}=a^{b-1}.....\mathbf{\text{1}}}\\ \text{maka}& \\ {\displaystyle \frac{a}{b}}& =a^{3b}...............\mathbf{\text{2}}\\ \mathbf{\text{1}}& ke\mathbf{\text{2}}\\ {\displaystyle \frac{a}{a^{b-1}}}& =a^{3b}\\ a^{2-b}& =a^{3b}\\ 2-b& =3b\\ -4b& =-2\\ b& ={\displaystyle \frac{1}{2}................\mathbf{\text{3}}}\\ \mathbf{\text{3}}& ke\mathbf{\text{1}}\\ a\left({\displaystyle \frac{1}{2}}\right)& =a^{\frac{1}{2}}\\ {\displaystyle \frac{1}{4}{a}^2}& =a\\ a^2-4a& =0\\ a(a-4)& =0\\ a=0& \text{atau}a=4\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Jika}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}=4,\text{maka}x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1,1}\\ \text{b}.& {\displaystyle 1,2}\\ \text{c}.& 1,3\\ \text{d}.& {\displaystyle 1,4}\\ \text{e}.& 1,5\\ \\ & & & (\mathbf{\text{SAT Test Math Level 2}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}}& =4\\ {\left(3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}\right)}^2& =4^2\\ 3^2.x^3& =4^2\\ x^3& ={\displaystyle \frac{4^2}{3^2}}\\ x^3& ={\displaystyle \frac{4^2}{3^2}\times \frac{3}{3}}\\ x^3& \le {\displaystyle \frac{4^2}{3^2}\times \frac{4}{3}}\\ x^3& \le \left({\displaystyle \frac{4^3}{3^3}}\right)\\ x^3& \le {\left({\displaystyle \frac{4}{3}}\right)}^3\\ x& \le {\displaystyle \frac{4}{3}}\\ x& \le 1,\overline{333}\\ x& \approx 1,3\end{array}\end{array}$

$\begin{array}{l}\\ 51.& \text{Hitunglah}\\ \\ & {\displaystyle \sqrt[8]{2207-{\displaystyle \frac{1}{2207-{\displaystyle \frac{1}{2207-{\displaystyle \frac{1}{2207-{\displaystyle \frac{1}{...}}}}}}}}}}\\ \\ & \text{nyatakan jawabannya dalam bentuk }{\displaystyle \frac{a\pm b\sqrt{c}}{d}}\\ & \text{dengan a, b, c, dan d bilangan-bilangan bulat}\end{array}$

Pembahasan:

$\begin{array}{rl}{x}^8& =2207-{\displaystyle \underset{x^8}{\underbrace{{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}}}\\ x^8& =2207-{\displaystyle \frac{1}{x^8}}\\ x^8+{\displaystyle \frac{1}{x^8}}& =2207\\ {(x^4+{\displaystyle \frac{1}{x^4}})}^2& =2207+2\\ (x^4+{\displaystyle \frac{1}{x^4}})& =\sqrt{2209}=47\end{array}$

$\begin{array}{rl}{x}^4+{\displaystyle \frac{1}{x^4}}& =47\\ {(x^2+{\displaystyle \frac{1}{x^2}})}^2& =47+2\\ x^2+{\displaystyle \frac{1}{x^2}}& =\sqrt{49}=7\\ {(x+{\displaystyle \frac{1}{x}})}^2& =7+2\\ x+{\displaystyle \frac{1}{x}}& =\sqrt{9}=3\\ x^2-3x+1& =0,\\ & \text{persamaan kuadrat dalam x,}\\ & \mathbf{\text{gunakan rumus abc}}\\ x_{1,2}=& {\displaystyle \frac{3\pm \sqrt{5}}{2}={\displaystyle \frac{3\pm 1\sqrt{5}}{2}={\displaystyle \frac{a\pm b\sqrt{c}}{d}}}}\\ & \mathbf{\text{Sehingga}},\{\begin{array}{ll}& a=3\\ & b=1\\ & c=5\\ & d=2\end{array}\end{array}$

DAFTAR PUSTAKA

1. Enung, S., Untung, W. 2009. Mandiri Matematika SMA Jilid I untuk Kelas X. Jakarta: ERLANGGA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
3. Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
4. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO.