Latihan Soal 4 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

$\begin{array}{l}\\ 31.& \text{Jika bentuk}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}\\ & \text{dinyatakan dalam pangkat positif}=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a^2}{a-b}}& & & \\ \\ \text{b}.& {\displaystyle \frac{a^2}{a-1}}\\ \\ \text{c}.& {\displaystyle \frac{b-a}{ab}}\\ \\ \text{d}.& {\displaystyle \frac{a^2}{b-a}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a-b}}\\ \\ & & & & (\mathbf{\text{SAT Test Math Level 2}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}& ={\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}}\\ & ={\displaystyle \frac{a^2}{b-a}}\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misalkan}A& =\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =3\\ \text{dikuadratkan}& \\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =9\\ x+3& =9\\ x& =9-3\\ x& =6\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Nilai}x\text{yang memenuhi}\\ & x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 7\sqrt[7]{7}\\ \text{b}.& 7\\ \text{c}.& 14\\ \text{d}.& 49\\ \text{e}.& \sqrt[3]{81}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x& =\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49x\\ x^2& =49\\ x& =\sqrt{49}\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai}x\text{yang memenuhi}\\ & x^{x^{x^{x^{x^{\cdots }}}}}=2020\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2020}\\ \text{b}.& \sqrt[2020]{2020}\\ \text{c}.& {2020}^{\sqrt{2020}}\\ \text{d}.& {\sqrt{2020}}^{\sqrt{2020}}\\ \text{e}.& \sqrt{2020\sqrt{2020}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^{x^{x^{x^{x^{\cdots }}}}}& =2020\\ x^{2020}& =2020\\ x& =\sqrt[2020]{2020}\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai dari}\\ & {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& \sqrt[3]{2}+1\\ \text{c}.& \sqrt[3]{2}-1\\ \text{d}.& \sqrt[3]{4}+1\\ \text{e}.& \sqrt[3]{4}-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}}\\ & ={\displaystyle \frac{{\left(\sqrt[3]{2}\right)}^2-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{2-1}}\\ & =\sqrt[3]{4}-1\end{array}\end{array}$

$\begin{array}{l}\\ 36.& \text{Nilai dari}\\ & {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\sqrt{2}-1\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \text{d}.& \sqrt{{\displaystyle \frac{5}{3}}}\\ \text{e}.& \sqrt{{\displaystyle \frac{2}{5}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-(\sqrt{2}-1)}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{3+\sqrt{5}}{\sqrt{2}}}\right)+\left({\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}\right)}{\sqrt{5}+1}+1-\sqrt{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}}{1+\sqrt{5}}+1-\sqrt{2}}\\ & ={\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}}\\ & =\sqrt{2}+1-\sqrt{2}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 37.& \text{Bentuk sederhana dari}\\ & \sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\ \\ & (\mathbf{\text{SIMAK UI 2012 Mat IPA}})\\ & \begin{array}{l}\\ \text{a}.& 2-\sqrt{2}\\ \text{b}.& 8-\sqrt{2}\\ \text{c}.& -2+\sqrt{2}\\ \text{d}.& 2+5\sqrt{2}\\ \text{e}.& 8+5\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{misalkan},\\ & \begin{array}{rl}x& =\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ & =(\sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}})\\ & =\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^2& =33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{(33+20\sqrt{2})(27-18\sqrt{2})}\\ & =60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ & =60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ & =60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ & =60+2\sqrt{2}-2(\sqrt{162}-\sqrt{9})\\ & =60+2\sqrt{2}-2(9\sqrt{2}-3)\\ x^2& =66-16\sqrt{2}\\ x& =\sqrt{66-2.8\sqrt{2}}\\ & =\sqrt{64+2-2\sqrt{64.2}}\\ & =\sqrt{64}-\sqrt{2}\\ & =8-\sqrt{2}\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Jika}{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}={\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12},}}\\ \\ & \text{maka}a+b+c=....\\ & (\mathbf{\text{UM UGM 2016 Mat Das}})\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}\\ & ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times {\displaystyle \frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}-\sqrt{5})}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-{\left(\sqrt{5}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}}\\ & \{\begin{array}{ll}a& =3\\ b& =2\\ c& =-1\end{array}\\ a+b+c& =3+2+(-1)\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Tunjukkan bahwa}\\ & \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots }}}}}=3\\ \\ & \text{Bukti}\\ & \begin{array}{rl}{x}^2& =x^2\\ x^2& =1+(x^2-1)\\ & =1+(x-1)(x+1)\\ & =1+(x-1)\sqrt{{(x+1)}^2}\\ & =1+(x-1)\sqrt{1+{(x+1)}^2-1}\\ & =1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & =1+(x-1)\sqrt{1+x(x+2)}\\ & =1+(x-1)\sqrt{1+x\sqrt{{(x+2)}^2}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+{(x+2)}^2-1}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{{(x+3)}^2}}}\\ x^2& =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}\\ x& =\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}}\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Jika terdapat hubungan berikut}\\ & \text{a}.2^p=3^q=6^r,\text{tunjukkan bahwa}pr+qr-pq=0\\ & \text{b}.2^x=3^{2y}=6^z,\text{tunjukkan bahwa }2xy-2yz-xz=0\\ & \text{c}.3^{15a}=5^{5b}={15}^{3c},\text{tunjukkan bahwa }5ab-bc-3ac=0\end{array}$

$\mathbf{\text{bukti}}$

Yang akan ditunjukkan adalah no. 40 yang poin c, yaitu:

$\begin{array}{rl}{3}^{15a}& =5^{5b}={15}^{3c}\{\begin{array}{ll}3=5^{\frac{5b}{15a}}& \\ 3^{\frac{15a}{5b}}=b& (a^b=c^d\to a=c^{\frac{d}{b}}\text{atau}{a}^{\frac{b}{d}}=c)\end{array}\\ 3^{15a}& ={15}^{3c}\\ 3^{15a}& =(3\times 5{)}^{3c}\\ 3^{15a}& =(3\times 3^{\frac{15a}{5b}}{)}^{3c}\\ 3^{15a}& =3^{3c+\frac{9c}{b}}\\ a^{f(x)}& =a^{g(x)}\\ f(x)& =g(x)\\ 15a& =3c+\frac{9ac}{b}\\ 15ab& =3bc+9ac\\ 5ab& =bc+3ac\\ 5ab-bc-3ac& =0◼\end{array}$