A.   \(3^{15} + 3^{15} + 3^{15}\)

\begin{equation*} \begin{split} & 3^{15} + 3^{15} + 3^{15} \\\\ & 3 \:.\: 3^{15} \\\\ & \bbox[5px, border: 2px solid magenta] {3^{16}} \end{split} \end{equation*}

B.   \(2^{25} + 2^{25} + 2^{25} + 2^{25}\)

\begin{equation*} \begin{split} & 2^{25} + 2^{25} + 2^{25} + 2^{25} \\\\ & 4 \:.\: 2^{25} \\\\ & 2^2 \:.\: 2^{25} \\\\ & \bbox[5px, border: 2px solid magenta] {2^{27}} \end{split} \end{equation*}

A.  \(\dfrac{a^4}{a^3 \:.\: b^2} \times \dfrac{a^7 \:.\: b^3}{a \:.\: b^6}\)

\begin{equation*} \begin{split} & \frac{a^4}{a^3 \:.\: b^2} \times \frac{a^7 \:.\: b^3}{a \:.\: b^6} \\\\ & \frac{a^4 \:.\: a^7}{a^3 \:.\: a} \times \frac{b^3}{b^2 \:.\: b^6} \\\\ & \frac{a^{11}}{a^4} \times \frac{b^3}{b^8} \\\\ & \frac{a^7}{1} \times \frac{1}{b^5} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{a^7}{b^5}} \end{split} \end{equation*}

B.   \(\left(\dfrac{x^5 \:.\: y}{x^3 \:.\: y^4}\right)^4 \div \left(\dfrac{x^4}{x \:.\: y^2}\right)^6\)

\begin{equation*} \begin{split} & \left(\frac{x^5 \:.\: y}{x^3  \:.\: y^4}\right)^4 \div \left(\frac{x^4}{x \:.\: y^2}\right)^6 \\\\ & \left(\frac{x^2}{y^3}\right)^4 \div \left(\frac{x^3}{y^2}\right)^6 \\\\ & \frac{x^8}{y^{12}} \div \frac{x^{18}}{y^{12}} \\\\ & \frac{x^8}{y^{12}} \times \frac{y^{12}}{x^{18}} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{1}{x^{10}}} \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{(x^{\frac{2}{3}} \:.\: y^{\frac{2}{3}})^{-2}}{(x^{\frac{1}{2}} \:.\: y^{\frac{1}{4}})^{-3}} \\\\ & \frac{x^{-\frac{4}{3}} \:.\: y^{-\frac{4}{3}}}{x^{-\frac{3}{2}} \:.\: y^{-\frac{3}{4}}} \\\\ & x^{-\frac{4}{3} + \frac{3}{2}} \:.\: y^{-\frac{4}{3} + \frac{3}{4}} \\\\ & x^{\frac16} \:.\: y^{-\frac{7}{12}} \\\\ & \frac{x^{\frac{1}{6}}}{y^{\frac{7}{12}}} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{\sqrt [6] {x}}{\sqrt [12] {y^7}}} \end{split} \end{equation*}

A.   \(\dfrac{x \:.\: y^{-1} - y \:.\: x^{-1}}{x^{-1} + y^{-1}}\)

\begin{equation*} \begin{split} & \frac{x \:.\: y^{-1} - y \:.\: x^{-1}}{x^{-1} + y^{-1}} \\\\ & \frac{\dfrac{x}{y} - \dfrac{y}{x}}{\dfrac{1}{x} + \dfrac{1}{y}} \quad {\color {blue} \times \: \frac{xy}{xy}} \\\\ & \frac{x^2 - y^2}{y + x} \\\\ & \frac{\cancel{(x + y)}(x - y)}{\cancel{y + x}} \\\\ & \bbox[5px, border: 2px solid magenta] {x - y} \end{split} \end{equation*}

B.   \(\dfrac{a \:.\: b^{-2} + a^{-2} \:.\: b}{a^{-2} - b^{-2}}\)

\begin{equation*} \begin{split} & \frac{a \:.\: b^{-2} + a^{-2} \:.\: b}{a^{-2} - b^{-2}} \\\\ & \frac{\dfrac{a}{b^2} + \dfrac{b}{a^2}}{\dfrac{1}{a^2} - \dfrac{1}{b^2}} \quad  {\color {blue} \times \: \frac{a^2 \: b^2}{a^2 \: b^2}} \\\\ & \frac{a^3 + b^3}{b^2 - a^2} \\\\ & \frac{\cancel{(a + b)}(a^2 + ab + b^2)}{\cancel{(b + a)}(b - a)} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{a^2 + ab + b^2}{b - a}} \end{split} \end{equation*}

A.   \(\dfrac{3^{n + 1} + 3^n}{3^{n + 2} + 3^{n - 1}}\)

\begin{equation*} \begin{split} & \frac{3^{n + 1} + 3^n}{3^{n + 2} + 3^{n - 1}} \\\\ & \frac{3^n \:.\: 3 + 3^n}{3^n \:.\: 3^2  + 3^n \:.\: 3^{- 1}} \\\\ & \frac{ \cancel{3^n} \:.\: (3 + 1)}{\cancel{3^n} \:.\: (3^2 + 3^{- 1})} \\\\ & \frac{4}{9 + \frac{1}{3}} \quad {\color {blue} \times \: \frac{3}{3}} \\\\ & \frac{12}{27 + 1} \\\\ & \frac{12}{28} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{3}{7}} \end{split} \end{equation*}

B.   \(\dfrac{(x^{a + 1})^b \:.\: x^{a + b}}{x^{a(b + 1)} \:.\: x^{2b}}\)

\begin{equation*} \begin{split} & \frac{(x^{a + 1})^b \:.\: x^{a + b}}{x^{a(b + 1)} \:.\: x^{2b}} \\\\ & \frac{x^{ab} \:.\: x^b \:.\: x^a \:.\: x^b}{x^{ab} \:.\: x^a \:.\: x^{2b}} \\\\ & \frac{x^{ab} \:.\: x^a \:.\: x^{2b}}{x^{ab} \:.\: x^a \:.\: x^{2b}} \\\\ & \bbox[5px, border: 2px solid magenta] {1} \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{\left(\dfrac{2}{5}\right)^{-2} \:.\: (32)^{\frac{3}{5}} - \left(\dfrac{1}{8}\right)^{-\frac{4}{3}}}{\left(\dfrac{8}{125}\right)^{-\frac{1}{3}} \:.\: \left(\dfrac{9}{4}\right)^{-\frac{1}{2}} \:.\: (27)^{\frac{2}{3}} + \left(\dfrac{1}{3}\right)^{-2}} \\\\ & \frac{\left(\dfrac{5}{2}\right)^{2} \:.\: (2^5)^{\frac{3}{5}} - \left(2^{-3}\right)^{-\frac{4}{3}}}{\left(\dfrac{5^3}{2^3}\right)^{\frac{1}{3}} \:.\: \left(\dfrac{2^2}{3^2}\right)^{\frac{1}{2}} \:.\: (3^3)^{\frac{2}{3}} + \left(3^{-1}\right)^{-2}} \\\\ & \frac{\dfrac{25}{4} \:.\: 2^3 - 2^4}{\dfrac{5}{2} \:.\: \dfrac{2}{3} \:.\: 3^2 + 3^2} \\\\ & \frac{50 - 16}{15 + 9} \\\\ & \frac{34}{24} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{17}{12}} \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{(0,25)^{-\frac{1}{2}} \:.\: (0,125)^{-\frac{2}{3}} + (0,2)^{-2}}{(0,5)^{-2} + (0,2)^{-1} + 0,375 + (0,25)^0} \\\\ & \frac{\left(\dfrac{1}{4}\right)^{-\frac{1}{2}} \:.\: \left(\dfrac{1}{8}\right)^{-\frac{2}{3}} + \left(\dfrac{1}{5}\right)^{-2}}{\left(\dfrac{1}{2}\right)^{-2} + \left(\dfrac{1}{5}\right)^{-1} + \dfrac{3}{8} + 1} \\\\ & \frac{(2^{-2})^{-\frac{1}{2}} \cdot (2^{-3})^{-\frac{2}{3}} + (5^{-1})^{-2}}{(2^{-1})^{-2} + (5^{-1})^{-1} + \dfrac{3}{8} + 1} \\\\ & \frac{2 \:.\: 4 + 25}{4 + 5 + \dfrac{3}{8} + 1} \\\\ & \frac{33}{10 + \dfrac{3}{8}} \quad {\color {blue} \times \: \frac{8}{8}} \\\\ & \frac{264}{80 + 3} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{264}{83}} \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{(0,111\dotso)^{2} + (6,25)^{\frac{1}{2}}}{(0,555\dotso)^2 + (2,25)^{\frac{1}{2}}} \\\\ & \frac{\left(\dfrac{1}{9}\right)^{2} + \left(6 \dfrac{1}{4}\right)^{\frac{1}{2}}}{\left(\dfrac{5}{9}\right)^2 + \left(2 \dfrac{1}{4}\right)^{\frac{1}{2}}} \\\\ & \frac{\left(\dfrac{1}{9}\right)^{2} + \left(\dfrac{25}{4}\right)^{\frac{1}{2}}}{\left(\dfrac{5}{9}\right)^2 + \left(\dfrac{9}{4}\right)^{\frac{1}{2}}} \\\\ & \frac{\dfrac{1}{81} + \dfrac{5}{2}}{\dfrac{25}{81} + \dfrac{3}{2}} \quad {\color {blue} \times \: \frac{81 \:.\: 2}{81 \:.\: 2}} \\\\ & \frac{2 + 405}{50 + 243} \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{407}{293}} \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{60^8 \:.\: 45^7 \:.\: 72^6}{50^6 \:.\: 48^7 \:.\: 75^8} \\\\ & \frac{(2^2 \:.\: 3 \:.\: 5)^8 \:.\: (3^2 \:.\: 5)^7 \:.\: (2^3 \:.\: 3^2)^6}{(2 \:.\: 5^2)^6 \:.\: (2^4 \:.\: 3)^7 \:.\: (3 \:.\: 5^2)^8} \\\\ & \frac{2^{16} \:.\: 3^8 \:.\: 5^8 \:.\: 3^{14} \:.\: 5^7 \:.\: 2^{18} \:.\: 3^{12}}{2^6 \:.\: 5^{12} \:.\: 2^{28} \:.\: 3^7 \:.\: 3^8 \:.\: 5^{16}} \\ \end{split} \end{equation*}

\begin{equation*} 2^{16 + 18 - 6 - 28} \:.\: 3^{8 + 14 + 12 - 7 - 8} \:.\: 5^{8 + 7 -12 - 16} \\ \end{equation*}

\begin{equation*} 2^{0} \:.\: 3^{19} \:.\: 5^{-13} \\ \end{equation*}

a = 0, b = 19 dan c = −13

(A)   \(25^{600}, \quad 81^{400}, \quad 256^{300}\)

\begin{equation*} \begin{split} 25^{600} & = (5^2)^{600} = 5^{1200} = (5^{3})^{400} = (125)^{400} \quad {\color {blue} \dotso \: (3)}\\\\ 81^{400} & = (3^4)^{400} = 3^{1600} = (3^{4})^{400} = (81)^{400} \quad {\color {blue} \dotso \: (2)}  \\\\ 256^{300} & = (2^8)^{300} = 2^{2400} = (2^{6})^{400} = (64)^{400} \quad  {\color {blue} \dotso \: (1)} \end{split} \end{equation*}

\(256^{300} <81^{400} < 25^{600} \)

A.   \(9^{x} + 9^{-x}\)

\begin{equation*} \begin{split} & 3^x + 3^{-x} = 5 \quad {\color {blue} \text{(kuadratkan kedua ruas)}}\\\\ & \left(3^x + 3^{-x}\right)^2 = 25 \\\\ & \left(3^x \right)^2 + 2 \:.\: 3^x \:.\: 3^{-x} + \left(3^{-x}\right)^2 = 25 \\\\ & 3^{2x} + 2 \:.\: 3^0 + 3^{-2x} = 25 \\\\ & 3^{2x} + 2 \:.\: 1 + 3^{-2x} = 25 \\\\ & 3^{2x} + 3^{-2x} = \bbox[5px, border: 2px solid magenta] {23} \end{split} \end{equation*}

A.   Jika \(2^{2x} + 2^{-2x} = 47\), tentukan nilai dari \(2^{3x} + 2^{-3x}\)

\begin{equation*} \begin{split} & \left(2^x + 2^{-x}\right)^2 = \left(2^x \right)^2 + 2 \:.\: 2^x \:.\: 2^{-x} + \left(2^{-x} \right)^2 \\\\ & \left(2^x + 2^{-x}\right)^2 = 2^{2x} + 2 \:.\: 2^0 + 2^{-2x} \\\\ & \left(2^x + 2^{-x}\right)^2 = 47 + 2 \\\\ & \left(2^x + 2^{-x}\right)^2 = 49 \\\\ & 2^x + 2^{-x} = 7 \end{split} \end{equation*}

\begin{equation*} \begin{split} & 2^x + 2^{-x} = 7 \quad {\color {blue} \text{(pangkat tiga kedua ruas)}}\\\\ & \left(2^x + 2^{-x}\right)^3 = 343 \\\\ & \left(2^x \right)^3 + 3 \:.\: \left(2^x \right)^2 \:.\: 2^{-x} + 3 \:.\: 2^x \:.\: \left( 2^{-x} \right)^2 + \left(2^{-x}\right)^3 = 343 \\\\ & 2^{3x} + 3 \:.\: 2^{2x} \:.\: 2^{-x} + 3 \:.\: 2^x \:.\: 2^{-2x} + 2^{-3x} = 343 \\\\ & 2^{3x} + 3 \:.\: 2^{x} + 3 \:.\: 2^{-x} + 2^{-3x} = 343 \\\\ & 2^{3x} + 3 \left( 2^{x} + 2^{-x} \right) + 2^{-3x} = 343 \\\\ & 2^{3x} + 3 \:.\: 7 + 2^{-3x} = 343 \\\\ & 2^{3x} + 21 + 2^{-3x} = 343 \\\\ & 2^{3x} + 2^{-3x} = \bbox[5px, border: 2px solid magenta] {322} \end{split} \end{equation*}

B.   Jika \(x^{2} + \dfrac {1}{x^2} = 14\), tentukan nilai dari \(x^{3} + \dfrac {1}{x^3}\)

\begin{equation*} \begin{split} & \left(x + \frac 1x \right)^2 = x^2 + 2 \:.\: x \:.\: \frac 1x + \left(\frac 1x \right)^2 \\\\ & \left(x + \frac 1x \right)^2 = x^2 + 2 + \frac {1}{x^2} \\\\ & \left(x + \frac 1x \right)^2 = 14 + 2 \\\\ & \left(x + \frac 1x \right)^2 = 16 \\\\ & x + \frac 1x = 4 \end{split} \end{equation*}

\begin{equation*} \begin{split} & x + \frac 1x = 4 \quad {\color {blue} \text{(pangkat tiga kedua ruas)}}\\\\ & x^3 + 3 \:.\: x^2 \:.\: \frac 1x + 3 \:.\: x \:.\: \left(\frac 1x \right)^2 + \left( \frac 1x \right)^3 = 64 \\\\ & x^3 + 3 \:.\: x + 3 \:.\: \frac 1x + \frac {1}{x^3} = 64 \\\\ & x^3 + 3 \left( x + \frac 1x \right) + \frac {1}{x^3} = 64 \\\\ & x^3 + 3 \:.\: 4 + \frac {1}{x^3} = 64 \\\\ & x^3 + 12 + \frac {1}{x^3} = 64 \\\\ & x^3 + \frac {1}{x^3} = \bbox[5px, border: 2px solid magenta] {52} \end{split} \end{equation*}