PAS MATEMATIKA BLOG

A. Sifat-Sifat Eksponen dan Logaritma

$\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Sifat yang berlaku pada}}\\\hline \textrm{Eksponens}&\textrm{Logaritma}\\\hline \displaystyle a^{n}=\underset{n\: \: faktor}{\underbrace{a\times a\times a\times \cdots \times a}}&^{a}\log b=c\: \Rightarrow \: a^{c}=b\\\hline \bullet \quad a^{p}\times a^{q}=a^{p+q}&\bullet \quad ^{a}\log x+\: ^{a}\log y=\: ^{a}\log xy\\\hline \bullet \quad a^{p}: a^{q}=a^{p-q}&\bullet \quad ^{a}\log x-\: ^{a}\log y=\: ^{a}\log \displaystyle \frac{x}{y}\\\hline \bullet \quad \left ( a^{p} \right )^{q}=a^{p.q}&\bullet \quad ^{a}\log x=\: \displaystyle \frac{^{m}\log x}{^{m}\log a}\\\hline \bullet \quad \displaystyle \sqrt[q]{a^{p}}=\displaystyle a^{ \left (\frac{p}{q} \right )}&\bullet \quad ^{a}\log b\: \times \: ^{b}\log c=\: ^{a}\log c\\\hline \bullet \quad \left ( a\times b \right )^{p}=a^{p}\: \times \: b^{p}&\bullet \quad ^{a^{m}}\log b^{n}=\displaystyle \frac{n}{m}\times \: ^{a}\log b\\\hline \bullet \quad \left ( \displaystyle \frac{a}{b} \right )^{p}=\displaystyle \frac{\displaystyle a^{p}}{\displaystyle b^{p}}&\bullet \quad \displaystyle a^{\: {^{a}}\log b}=b\\\hline \bullet \quad a^{-p}=\displaystyle \frac{1}{\displaystyle a^{p}}&\bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{b}\log a}\\\hline \bullet \quad a^{0}=1,\: \: \: \: \: a\neq 0&\bullet \quad ^{a}\log 1=0\\\hline \bullet \quad a^{1}=1&\bullet \quad ^a\log a=1\\\hline \begin{cases} a,b\: \in \mathbb{R} \\ p,q\: \in \mathbb{Q} \end{cases}&\begin{cases} a\neq 0 & a>0\: \: (\textrm{bilangan pokok}) \\ x,y>0 & (\textrm{numerus}) \end{cases}\\\hline \end{array}$

B. Persamaan Eksponen dan Logaritma

B.1 Persamaan Eksponen

$\begin{array}{|l|l|l|}\hline \textrm{No}&\textrm{Bentuk}&\textrm{Syarat}\\\hline 1.&a^{f(x)}=1&a\neq 0,\quad \textrm{maka}\: \: f(x)=0\\\hline 2.&a^{f(x)}=a^{p}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: f(x)=p\\\hline 3.&a^{f(x)}=a^{g(x)}&a>0,\: \: a\neq 1,\quad \textrm{maka}\: \: f(x)=g(x)\\\hline 4.&a^{f(x)}=b^{f(x)}&a\neq 0,\: b\neq 0\: ,\quad \textrm{maka}\: \: f(x)=0\\\hline 5.&f(x)^{g(x)}=1&\begin{cases} f(x)=1 & \\ g(x)=0, & \textrm{jika}\: \: f(x)\neq 0 \\ f(x)=-1, & \textrm{jika}\: \: g(x)=\: \textrm{genap} \end{cases}\\\hline 6.&f(x)^{g(x)}=f(x)^{h(x)}&\begin{cases} (i).\quad g(x)=h(x)& \\ (ii).\quad f(x)=1& \\ (iii).\quad f(x)=0,&g(x)>0,\: \: h(x)>0 \\ (iv).\quad f(x)=-1,&g(x)\: \textrm{dan}\: h(x)\: \: \\ &\textrm{keduanya ganjil atau genap} \end{cases}\\\hline 7.&g(x)^{f(x)}=h(x)^{f(x)}&\begin{cases} (i).\quad g(x) =h(x)& \\ (ii).\quad f(x)=0, & g(x)\neq 0,\: h(x)\neq 0 \end{cases}\\\hline 8.&A\left ( a^{f(x)} \right )^{2}+B\left ( a^{f(x)} \right )+C=0&a>0,\: \: a\neq 1\\\hline \end{array}$

B.2 Persamaan Logaritma

$\begin{array}{|l|l|l|}\hline \textrm{No}&\textrm{Bentuk}&\textrm{Syarat}\\\hline 1.&^a\log f(x)=0&f(x)>0,a>0,a\neq 0,\quad \textrm{maka}\: \: f(x)=1\\\hline 2.&^a\log f(x)=\: ^a\log p&a>0, a\neq 1, f(x)>0,p>0\quad \textrm{maka}\: \: f(x)=p\\\hline 3.&^a\log {f(x)}=\: ^a\log {g(x)}&a>0, a\neq 1,f(x)>,g(x)>0 \quad \textrm{maka}\: \: f(x)=g(x)\\\hline 4.&^a\log {f(x)}=\: ^b\log {f(x)}&a>0,b>0,a\neq 1, b\neq 1,f(x)>0,g(x)>0\\ && \textrm{maka}\: \: f(x)=0\\\hline 5.&^{h(x)}\log f(x)=\: ^{h(x)}\log g(x)&h(x)>0,h(x)\neq 1,f(x)>0,g(x)>0\\ && \textrm{maka}\: \: f(x)=g(x)\\\hline 6.&A(^a\log ^{2}f(x))+B(^a\log f(x))+C=0&\textrm{arahkan ke persamaan kuadrat}\\\hline 7.&a^{f(x)}=b^{g(x)}&\textrm{gunakan aturan logaritma}\\\hline \end{array}$

C. Pertidaksamaan Eksponen dan Logaritma

C. 1 Pertidaksamaan Eksponen

$\begin{array}{|l|l|}\hline a>1&0<a<1\\\hline a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\leq g(x)&a^{f(x)}\leq a^{g(x)}\Rightarrow f(x)\geq g(x)\\\hline a^{f(x)}< a^{g(x)}\Rightarrow f(x)< g(x)&a^{f(x)}< a^{g(x)}\Rightarrow f(x)> g(x)\\\hline a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\geq g(x)&a^{f(x)}\geq a^{g(x)}\Rightarrow f(x)\leq g(x)\\\hline a^{f(x)}> a^{g(x)}\Rightarrow f(x)> g(x)&a^{f(x)}> a^{g(x)}\Rightarrow f(x)< g(x)\\\hline \end{array}$

C. 2 Pertidaksamaan Logaritma $\left (f(x)>0\: \: \textrm{dan}\: \: g(x)>0 \right )$

$\begin{array}{|l|l|}\hline a>1&0<a<1\\\hline ^a\log f(x)\leq \: ^a\log g(x)\Rightarrow f(x)\leq g(x)&^a\log f(x)\leq \: ^a\log g(x)\Rightarrow f(x)\geq g(x)\\\hline ^a\log f(x)< \: ^a\log g(x)\Rightarrow f(x)< g(x)&^a\log f(x)< \: ^a\log g(x)\Rightarrow f(x)> g(x)\\\hline ^a\log f(x)\geq \: ^a\log g(x)\Rightarrow f(x)\geq g(x)&^a\log f(x)\geq \: ^a\log g(x)\Rightarrow f(x)\leq g(x)\\\hline ^a\log f(x)> \: ^a\log g(x)\Rightarrow f(x)> g(x)&^a\log f(x)> \: ^a\log g(x)\Rightarrow f(x)< g(x)\\\hline \end{array}$ .

D. Grafik Fungsi Eksponen dan Logaritma

D.1 Grafik fungsi Eksponen

D.2 Grafik fungsi Logaritma

$\LARGE{\fbox{\LARGE{\fbox{CONTOH SOAL}}}}$ .

$\begin{array}{ll}\\ \fbox{1}.&\textrm{Tentukanlah nilai dari bilangan-bilangan berikut ini}! \end{array}\\ \begin{array}{llllllll}\\ .\quad\quad &a.&27^{\frac{1}{3}}&k.&\left ( \displaystyle \frac{2^{3}.3^{-2}}{2^{-5}.3} \right )^{\displaystyle \frac{1}{2}}&u.&\displaystyle \frac{\left ( a^{2}.b^{-1} \right )^{\frac{1}{2}}\sqrt{a^{6}.b^{\frac{5}{3}}.c^{-2}}}{\left ( a^{3}.b^{-5}.c^{-3} \right )^{\frac{1}{3}}}\\ &b.&32^{^{\frac{2}{5}}}&l.&\displaystyle \frac{\sqrt{2}.\sqrt[3]{8}}{(2)^{\frac{1}{3}}.\sqrt[4]{16^{2}}}&v.&\displaystyle \frac{x^{2}.y^{7}}{x^{3}.y^{5}}\\ &c.&\left ( \displaystyle \frac{9}{16} \right )^{\displaystyle \frac{3}{2}}&m.&\left ( \displaystyle \frac{\sqrt{2}.2\sqrt{6}}{\sqrt{3}.\sqrt[3]{9}} \right )^{\displaystyle \frac{1}{2}}&w.&\left ( \displaystyle \frac{2x^{3}}{y^{2}}:\frac{4x^{6}}{4y^{5}} \right ).\displaystyle \frac{3x^{2}-2y}{3y}\\ &d.&\left ( \displaystyle \frac{1}{125} \right )^{-\frac{1}{3}}&n.&\displaystyle \frac{2.3^{-\frac{1}{2}}-2+3.2^{-\frac{1}{2}}}{2.3^{-\frac{1}{2}}-3.2^{-\frac{1}{2}}}&x.&\left (\sqrt[3]{x^{2}.yz^{3}} \right ).x^{-1}.y^{-2}\\ &e.&\left ( \displaystyle \frac{2}{3} \right )^{\displaystyle \frac{2}{3}}.\left ( \displaystyle \frac{3}{2} \right )^{-\displaystyle \frac{1}{3}}&o.&-3\sqrt{6}+4\sqrt{3}-2\sqrt{81}&y.&\displaystyle \frac{\sqrt{x}.\sqrt{x^{2}y^{3}}.\sqrt{xy^{2}}}{\sqrt[4]{x}.\sqrt[3]{y}}\\ &f.&\left ( \displaystyle \frac{1}{5^{3}} \right )^{-1}.\left ( \displaystyle \frac{1}{5^{2}} \right )^{2}&p.&\sqrt{250}-\sqrt{50}+15\sqrt{2}&z.&\displaystyle \frac{\left ( x^{2} \right )^{3}}{x^{4}}:\left ( \frac{x^{3}}{\left ( x^{3} \right )^{2}} \right )^{-2} \end{array}$

$\begin{array}{llllllll}\\ .\quad\quad&g.&\displaystyle \frac{\sqrt{3}.\sqrt{15}}{\sqrt{5}}\quad\qquad \: \: &q.&\sqrt[2]{75}-\sqrt[4]{27}+\sqrt[3]{128}\\ &h.&\displaystyle \frac{2\sqrt{3}.\sqrt{24}}{\sqrt{7}.3\sqrt{14}}&r.&\displaystyle \frac{5\sqrt{5}+2\sqrt{5}}{5-3\sqrt{5}}\\ &i&\displaystyle \frac{\sqrt{5}.\sqrt[2]{2^{3}}}{2.\sqrt[3]{3}}&s.&\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots }}}}}}\\ &j.&\displaystyle \frac{\sqrt{3}.\sqrt[3]{2}}{\left ( \frac{4}{9} \right )^{3}.\left ( \frac{2}{3} \right )^{-2}}&t.&\left ( 4^{\frac{1}{2}} \right )^{\frac{1}{2}}\left ( 2^{-2} \right )^{-2}.\sqrt[3]{0,125}.\left ( 0,25 \right ).\displaystyle \frac{1}{2}\sqrt{\frac{1}{2}} \end{array}$

Pembahasan:

$\begin{array}{|l|l|l|l|}\hline \begin{aligned}a.\quad 27^{\frac{1}{3}}&=\left ( 3^{3} \right )^{\frac{1}{3}}\\ &=3\\ &\\ & \end{aligned}&\begin{aligned}b.\quad 32^{\frac{2}{5}}&=\left ( 2^{5} \right )^{\frac{2}{5}}\\ &=2^{2}=4\\ &\\ & \end{aligned}&\begin{aligned}c.\quad \left ( \displaystyle \frac{9}{16} \right )^{\frac{3}{2}}&=\left (\left ( \displaystyle \frac{3}{4} \right )^{2} \right )^{\frac{3}{2}}\\ &=\left ( \displaystyle \frac{3}{4} \right )^{3}=\frac{27}{64} \end{aligned}&\begin{aligned}d.\quad \left ( \displaystyle \frac{1}{125} \right )^{-\frac{1}{3}}&=\left ( 5^{-3} \right )^{-\frac{1}{3}}\\ &=5\\ & \end{aligned}\\\hline \begin{aligned}e.\quad &\left ( \frac{2}{3} \right )^{\frac{2}{3}}.\left ( \frac{3}{2} \right )^{-\frac{1}{3}}\\ &=\left ( \frac{2}{3} \right )^{\frac{2}{3}}.\left ( \frac{2}{3} \right )^{\frac{1}{3}}\\ &=\left ( \frac{2}{3} \right )^{\frac{2}{3}+\frac{1}{3}}\\ &=\frac{2}{3} \end{aligned}&\begin{aligned}h.\quad &\displaystyle \frac{2\sqrt{3}.\sqrt{24}}{\sqrt{7}.3\sqrt{14}}\\ &=\displaystyle \frac{2\sqrt{3}.\sqrt{4}.\sqrt{2}.\sqrt{3}}{\sqrt{7}.3.\sqrt{2}.\sqrt{7}}\\ &=\frac{4.\sqrt{2}.3}{7.\sqrt{2}.3}\\ &=\frac{4}{7} \end{aligned}&\begin{aligned}u.\quad &\displaystyle \frac{\left ( a^{2}b^{-1} \right )^{\frac{1}{2}}.\sqrt{a^{6}.b^{\frac{5}{3}}.c^{-2}}}{\left ( a^{3} \right )^{\frac{1}{3}}}\\ &=\displaystyle \frac{a.b^{-\frac{1}{2}}.a^{\frac{6}{2}}.b^{\frac{1}{2}.\frac{5}{3}}.c^{-\frac{2}{2}}}{a^{\frac{3}{3}}.b^{-\frac{5}{3}}.c^{-\frac{3}{3}}}\\ &=a^{3}.b^{-\frac{1}{2}+\frac{5}{6}+\frac{5}{3}}\\ &=a^{3}.b^{\frac{-3+5+10}{6}}=a^{3}.b^{2} \end{aligned}&\begin{aligned}&\textrm{Untuk Soal yang belum}\\ &\textrm{dibahas silahkan}\\ &\textrm{dikerjakan sendiri}\\ &\textrm{sebagai latihan} \end{aligned}\\\hline \end{array}$

$\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-1}=1\\ &\textrm{b}.\quad 4^{x^{2}+3x-10}=1\\ &\textrm{c}.\quad 5^{3x^{2}+2x-1}=1\\ &\textrm{d}.\quad (9)^{2x+\frac{1}{2}}.\left ( \displaystyle \frac{1}{27} \right )^{3x^{2}+2}=1\\ &\textrm{e}.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-3}.(8)^{3x+1}=1\\ &\textrm{f}.\quad \displaystyle \frac{\sqrt[3]{\left ( 0,0008 \right )^{7-2x}}}{\left ( 0,2 \right )^{-4x+5}}=1\: \: ...(\textbf{SPMB 2005})\end{array}$

Pembahasan:

$\begin{array}{|l|l|l|}\hline \begin{aligned}a.\quad 3^{2x-1}&=1\\ 3^{2x-1}&=3^{0}\\ 2x-1&=0\\ 2x&=1\\ x&=\displaystyle \frac{1}{2}\\ \textrm{HP}=&\left \{ \displaystyle \frac{1}{2} \right \}\\ &\\ &\\ & \end{aligned}&\begin{aligned}e.\quad \left ( \displaystyle \frac{1}{2} \right )^{2x-3}.(8)^{3x+1}&=1\\ (2^{3-2x}).(2^{3})^{3x+1}&=2^{0}\\ 2^{3-2x+9x+3}&=2^{0}\\ 7x+6&=0\\ 7x&=-6\\ x&=-\displaystyle \frac{6}{7}\\ \textrm{HP}=\left \{ -\displaystyle \frac{6}{7} \right \}&\\ & \end{aligned}&\begin{aligned}f.\quad \displaystyle \frac{\sqrt[3]{\left ( 0,0008 \right )^{7-2x}}}{\left ( 0,2 \right )^{-4x+5}}&=1\: \: ...(\textbf{SPMB 2005})\\ \displaystyle \frac{((0,2)^{3})^{\frac{7-2x}{3}}}{(0,2)^{-4x+5}}&=(0,2)^{0}\\ (0,2)^{7-2x-(-4x+5)}&=(0,2)^{0}\\ 7-2x+4x-5&=0\\ 2x+2&=0\\ 2&=-2\\ x&=-1\\ \textrm{HP}=\left \{ -1 \right \}& \end{aligned}\\\hline \end{array}$

$\begin{array}{ll}\\ \fbox{3}.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad 3^{2x-5}=3^{5}\\ &\textrm{b}.\quad 2^{2x+3}.\left (\displaystyle \frac{1}{8} \right )^{x-3}=\displaystyle \frac{1}{64}\\ &\textrm{c}.\quad (2)^{5x^{2}-3x}.\left ( \displaystyle \frac{1}{32} \right )^{5}=32^{5}\\ &\textrm{d}.\quad \left ( \displaystyle \frac{1}{9} \right )^{-x^{2}}.\left ( 3^{2} \right )^{3x-3}=\displaystyle 9^{3^{2}}\\ &\textrm{e}.\quad (4)^{-2x^{2}+3x}.\left ( \displaystyle \frac{1}{4} \right )^{3}=2^{-2}.\left ( \displaystyle \frac{1}{8} \right )^{-2}\\ &\textrm{f}.\quad \displaystyle \frac{27}{3^{2x-1}}=81^{-0,125}\: \: ...(\textbf{SPMB 2004})\end{array}$

Pembahasan:

$\begin{array}{|l|l|}\hline \begin{aligned}a.\quad 3^{2x-5}&=3^{5}\\ a^{f(x)}&=a^{p}\\ f(x)&=p\\ 2x-5&=5\\ 2x&=10\\ x&=5\\ \textrm{HP}=&\left \{ 5 \right \}\\ &\\ &\\ &\\ &\\ &\end{aligned}&\begin{aligned}f.\quad \displaystyle \frac{27}{3^{2x-1}}&=81^{-0,125}\cdots (\textbf{SPMB 2004})\\ \displaystyle \frac{3^{3}}{3^{2x-1}}&=(3^{4})^{-\frac{1}{8}}\\ 3^{3-(2x-1)}&=3^{-\frac{4}{8}}\\ 3-(2x-1)&=-\frac{4}{8}\\ 4-2x&=-\frac{1}{2}\\ 8-4x&=-1\\ -4x&=-1-8\\ x&=\frac{9}{4}\\ \textrm{HP}=&\left \{ \frac{9}{4} \right \}\end{aligned}\\\hline \end{array}$

$\begin{array}{ll}\\ \fbox{4}.&\textrm{Tentukanlah himpunan penyelesaian (HP) dari}\\ &\textrm{a}.\quad \sqrt{3^{2x+1}}=9^{x-2}\\ &\textrm{b}.\quad \left ( \displaystyle \frac{1}{3} \right )^{2x-3}.3^{x+5}=\left ( \displaystyle \frac{1}{27} \right )^{2x-10}\\ &\textrm{c}.\quad (125)^{(x^{2}-3x-4)}=(5)^{(x^{2}-2x-3)}\\ &\textrm{d}.\quad (2^{2})^{\displaystyle \sqrt{x^{3}+2x^{2}-3x-6}}=\left ( \displaystyle \frac{1}{2} \right )^{-\displaystyle \sqrt{4x^{2}+4x-8}}\\ &\textrm{e}.\quad (4)^{(x^{2}+2x-1)}.\left ( \displaystyle \frac{1}{8} \right )^{3x-4}=\left ( \displaystyle \frac{1}{2} \right )^{2x}\\ &\textrm{f}.\quad (\sqrt{2})^{4x^{2}-8x+12}.\left ( \sqrt[3]{8} \right )^{3x+5}=\left ( \displaystyle \frac{1}{4} \right )^{4x-3}.(2)^{6x+5}\\ &\textrm{g}.\quad 9^{x^{2}-3x+1}+9^{-3x+x^{2}}=20-10\left ( 3^{x^{2}-3x} \right )\: \: ...(\textbf{SPMB 2004})\end{array}$

Pembahasan:

$\begin{array}{|l|l|l|}\hline \begin{aligned}a.\quad \sqrt{3^{2x+1}}&=9^{x-2}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ \displaystyle (3)^{\frac{2x+1}{2}}&=(3^{2})^{x-2}\\ \displaystyle \frac{2x+1}{2}&=2(x-2)\\ 2x+1&=4x-8\\ -2x&=-9\\ x&=\displaystyle \frac{9}{2}\\ \textrm{HP}=&\left \{ \frac{9}{2} \right \} \end{aligned}&\begin{aligned}b.\quad \left ( \displaystyle \frac{1}{3} \right )^{2x-3}.3^{x+5}&=\left ( \displaystyle \frac{1}{27} \right )^{2x-10}\\ (3^{-1})^{(2x-3)}.3^{x+5}&=(3^{-3})^{(2x-10)}\\ 3^{(-2x+3)+(x+5)}&=3^{10-2x}\\ -2x+3+x+5&=10-2x\\ x&=10-8\\ x&=2\\ \textrm{HP}=\left \{ 2 \right \}&\\ &\\ &\\ & \end{aligned}&\begin{aligned}c.\quad (125)^{x^{2}-3x-4}&=(5)^{x^{2}-2x-3}\\ (5^{3})^{x^{2}-3x-4}&=(5)^{x^{2}-2x-3}\\ 3x^{2}-9x-12&=x^{2}-2x-3\\ 2x^{2}-7x-9&=0\\ (x+1)(2x-9)&=0\\ x=-1\: \: \textrm{V}\: \: x=\frac{9}{2}\\ \textrm{HP}=&\left \{ -1,\frac{9}{2} \right \}\\ &\\ &\\ & \end{aligned}\\\hline \multicolumn{3}{|l|}{\begin{aligned}g.\qquad\quad\quad\quad 9^{x^{2}-3x+1}+9^{-3x+x^{2}}&=20-10\left ( 3^{x^{2}-3x} \right )\\ \textrm{misalkan}\: \: 3^{x^{2}-3x}&=p\\ 9^{x^{2}-3x}.9+9^{x^{2}-3x}&=20-10\left ( 3^{x^{2}-3x} \right )\\ 9.\left ( 3^{2.(x^{2}-3x)} \right )+\left ( 3^{2.(x^{2}-3x)} \right )&=20-10\left ( 3^{x^{2}-3x} \right )\\ 10.\left ( 3^{x^{2}-3x} \right )^{2}&=20-10\left ( 3^{x^{2}-3x} \right )\\ 10p^{2}&=20-10p\\ p^{2}&=2-p\\ p^{2}+p-2&=0\\ (p+2)(p-1)&=0\\ p=-2\: (\textrm{tdk mungkin})\: \: \textrm{V}\: \: p=1&\\ \textrm{sehingga}\quad p=3^{x^{2}-3x}&=1\\ 3^{x^{2}-3x}&=3^{0}\\ x^{2}-3x&=0\\ x(x-3)&=0\\ x=1\: \: \textrm{V}\: \: x&=3\\ \textrm{HP}=&\left \{ 1,3 \right \} \end{aligned}}\\\hline \end{array}$

Sumber Referensi

Kuntarti, Sulistiyono dan Sri Kurnianingsih. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
Kuntarti, Sulistiyono dan Sri Kurnianingsih. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: esis.
Soetiyono, Kamta Agus Sajaka, Sigit suprijanto, Marwanta, Suwarsini Murniati, dan Herynugroho. 2007. Matematika Interaktif 3B Sekolah Menengah Atas Kelas XII Program Ilmu Pengetahuan Alam. Jakarta: Yudistira.
Tim IGMP Matematika SMA. ….. . Tabloid Matematika Kurikulum 2006. Semarang: CV. Sarana Ilmu.
Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Saya pernah lihat soal seperti berikut

Soal tersebut meminta berapa besar tinggi H seperti ilustrasi gambar tersebut di atas dengan x = 8 cm dan y = 6 cm ?

Saya lihat soal ini di try out SMP, wah, mula-mula ada bingungnya juga. Pikir saya, saya mau mengerjakan dengan cara bagaimana? maklum kemampuan saya biasa-biasa aja. Iseng-iseng saya kerjakan soal tersebut dengan bantuan rekayasa diagram kartesius, saya menyebutnya demikianlah.. Saya yakin pembaca yang budiman ada yang gak setuju atau mungkin setuju dengan saya karena terlalu ribet.

Sukur-sukur di antara Anda para pembaca yang budiman ada yang sudi memberikan solusi dengan konsep kesebangunan atau apalah supaya kita tercerahkan.

Ok lanjut aja

Misalkan yang saya maksudkan di atas (bantuan diagram kartesius) seperti berikut:

Selanjutnya kita buat persamaan garisnya yaitu anggap saja garis yang melalui titik $O(0,0)$ adalah $L_{1}$ dan garis yang satunya kita sebut sebagai $L_{2}$.

Karena persamaan garis $L_{1}$ melalui titik $O(0,0)$ dan $(c,b)$, maka persamaan garisnya adalah

$\begin{aligned}\displaystyle \frac{y-0}{b-0}&=\frac{x-0}{c-0}\\ \Leftrightarrow y&=\color{red}\displaystyle \frac{b}{c}x\color{black}\\ \Leftrightarrow x&=\color{blue}\displaystyle \frac{c}{b}y\: \color{black}................(1) \end{aligned}$.

Untuk persamaan garis $L_{2}$ yang melalui titik $(c,0)$ dan $(0,b)$, persamaan garisnya adalah

$\begin{aligned}\frac{(y-0)}{(a-0)}&=\frac{(x-c)}{(0-c)}\\ y&=\color{blue}-\frac{a}{c}(x-c)\: \color{black}................(2) \end{aligned}$.

Dari persamaan 1) dan 2) kita mendapatkan

$\begin{aligned}y&=-\frac{a}{c}(x-c)\\\\ \Leftrightarrow\: \: & y=-\frac{a}{c}(\frac{c}{b}y-c)\\\\ \Leftrightarrow\: \: & y=-\frac{a}{b}y+a\\ \Leftrightarrow\: \: & y+\displaystyle \frac{a}{b}y=a\Leftrightarrow \displaystyle \frac{a+b}{b}y=a\\\\ \therefore &\quad y=\color{red}\frac{ab}{a+b} \end{aligned}$.

Jadi titnggi H adalah sebesar

$\textrm{H}=\displaystyle \frac{xy}{x+y}$.

Sehingga apabila x = 8 cm dan y = 6 cm , maka tinggi H adalah

$\textrm{H}=\displaystyle \frac{8.6}{8+6}=\frac{48}{14}=\frac{24}{7}$.

Sebagai catatannya adalah ternyata harga M tidak muncul dalam formula tersebut , unik memang. Tapi begitulah adanya.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ &(\textbf{OSN SD 2009})\\ &\textrm{Pada gambar berikut, diketahui}\: \: AB=3\: \: cm\\ &\textrm{dan}\: \: CD=4\: \: cm.\: \textrm{Sisi}\: \: AB,EF,\: \textrm{dan}\: \: CD\\ &\textrm{masing-masing tegak lurus}\: \: AC.\: \textrm{Tentukan}\\ &\textrm{panjang}\: \: EF?\\\\ \end{array}$.

$.\qquad\begin{aligned}\textrm{Dengan}&\textrm{rumus di atas akan dengan}\\ \textrm{mudah }&\textrm{kita tentukan panjangnya, yaitu}:\\ \quad EF&=\color{red}\frac{AB\times CD}{AB+CD}\\ &=\displaystyle \frac{3\times 4}{3+4}\\ &=\displaystyle \frac{12}{7}\: \: cm\\\\ \textrm{Jadi,}\: \textrm{ti}&\textrm{nggi}\: \: EF=\color{blue}\displaystyle \frac{12}{7}\: \: \color{black}cm \end{aligned}$.

PAS MATEMATIKA BLOG

Eksponen dan Logaritma

Keunikan Susunan Beberapa Bilangan

Menentukan Tinggi dari Perpotongan Dua Sinar

Bismillah mulai