Answer \( \color{red} \sqrt[\Large 5]{\left(\dfrac{45}{8}\right)^{3}}\)

Answer \( \color{red} \sqrt[\Large 9]{\left( 253 \right)^7}\)

Solution

\[ \begin{aligned} & = \left(\dfrac{-5}{81}\right)^{\tfrac{-4}{7}} \\ \\ & = \left(\dfrac{-81}{5}\right)^{\tfrac{4}{7}} \\ \\ & = \sqrt[7]{\left(\frac{-81}{5}\right)^{4}} \end{aligned} \]

Answer \( \color{red} \sqrt[\Large 7]{\left(\dfrac{-81}{5}\right)^{4}} \)

Answer \( \color{red} \left(\dfrac{-5}{11}\right)^{\tfrac{7}{9}}\)

Solution

\[ \begin{aligned} & = \sqrt[\Large 3]{\left(\frac{4}{9}\right)^{-2}} \\ \\ & = \left(\frac{4}{9}\right)^{\tfrac{-2}{3}} \\ \\ & = \left(\frac{9}{4}\right)^{\tfrac{2}{3}} \end{aligned} \]

Answer \( \color{red} \left(\dfrac{9}{4}\right)^{\tfrac{2}{3}} \)

Answer \( \color{red} (212)^{\frac{5}{7}}\)

Solution

\[ \begin{array}{c|c} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \] \[ \begin{aligned} (-3)^6 & = 729 \\ \end{aligned} \]

Answer The power is \( \color{red}6\)

Solution

\[ \begin{aligned} 2^4 + 3^2 &= 5^x \\ 16 + 9 &= 5^x \\ 25 &= 5^x \\ 5^2 &= 5^x \\ \text{Bases are the same, } & \text{we equate powers.} \\ \implies x & = 2 \end{aligned} \]

Answer \(x = \color{red}2\)

Solution

\[ \begin{aligned} & = \left(4^0 + 4^{-1}\right)\times 2^2 \\ \\ & = \left(1 + \frac{1}{4} \right)\times 4 \\ \\ & = \left(\frac{4 + 1}{4} \right)\times 4 \\ \\ & = \frac{5}{\cancel4_1} \times \cancel4^1 \\ \\ & = 5 \end{aligned} \]

Answer \( \color{red} 5 \)

Solution

\[ \begin{aligned} & = (-4)^5\div(-4)^8 \\[5pt] & = (-4)^{5-8} \\[5pt] & = (-4)^{-3} \\[5pt] & = \frac{-1}{4^{3}} \\[5pt] & = \frac{-1}{64} \end{aligned} \]

Answer \(\color{red} \dfrac{-1}{64}\)

Solution

\[ \begin{aligned} & =a^2\times a^3\times a^{-5} \\[5pt] & =a^{2 + 3 + (-5)} \\[5pt] & =a^{5 - 5} \\[5pt] & =a^0 \\[5pt] & = 1 \end{aligned} \]

Answer \(\color{red}1\)

Solution

\[ \begin{aligned} & =\bigl(\sqrt4\bigr)^{-3} \\[5pt] & = 2^{-3} \\[5pt] & = \frac{1}{2^{3}} \\[5pt] & = \frac{1}{8} \end{aligned} \]

Answer \(\color{red}\dfrac18\)

Solution

\[ \begin{aligned} & = \left[\left(\frac{1}{2}\right)^{-1}+\left(\frac{1}{3}\right)^{-1}\right]^{-1} \\[5pt] & = (2 + 3)^{-1} \\[5pt] & = 5^{-1} \\[5pt] & = \frac{1}{5} \end{aligned} \]

Answer \(\color{red}\dfrac15\)

Solution

\[ \begin{aligned} & = \left(3^{-7}\div3^{-9}\right)\times3^{-4} \\[5pt] & = \left[3^{-7 - (-9)}\right]\times3^{-4} \\[5pt] & = \left[3^{-7 + 9}\right]\times3^{-4} \\[5pt] & = 3^2 \times 3^{-4} \\[5pt] & = 3^{2+(-4)} \\[5pt] & = 3^{2-4} \\[5pt] & = 3^{-2} \\[5pt] & = \frac{1}{3^2} \\[5pt] & = \frac{1}{9} \end{aligned} \]

Answer \(\color{red}\dfrac19\)

Solution

\[ \begin{aligned} (-5)^{x+1}\times(-5)^5 & =(-5)^7 \\[5pt] (-5)^{x+1+5} & =(-5)^7 \\[5pt] (-5)^{6+x} & =(-5)^7 \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ 6+x & = 7 \\[5pt] x & = 7-6 \\[5pt] x & = 1 \\[5pt] \end{aligned} \]

Answer \( x = \color{red}1 \)

Solution

\[ \begin{aligned} (81)^{-4}\div(729)^{2-x} &= 9^{4x} \\[5pt] (3^4)^{-4}\div(3^6)^{2-x} &= (3^2)^{4x} \\[5pt] 3^{-16}\div3^{6(2-x)} &= 3^{8x} \\[5pt] 3^{-16}\div3^{12-6x} &= 3^{8x} \\[5pt] 3^{-16-(12-6x)} &= 3^{8x} \\[5pt] 3^{-16-12+6x} &= 3^{8x} \\[5pt] 3^{6x-28} &= 3^{8x} \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ 6x - 28 &= 8x \\[5pt] -28 &= 8x - 6x \\[5pt] -28 &= 2x \\[5pt] x &= \dfrac{-28}{2} \\[5pt] x &= -14 \end{aligned} \]

Answer \( x = \color{red}{-14} \)

Solution

\[ \begin{aligned} \left(\dfrac{11}{9}\right)^3 \times \left(\dfrac{9}{11}\right)^6 &= \left(\dfrac{11}{9}\right)^{2x-1}\\[5pt] \left(\dfrac{11}{9}\right)^3 \times \left(\dfrac{11}{9}\right)^{-6} &= \left(\dfrac{11}{9}\right)^{2x-1}\\[5pt] \left(\dfrac{11}{9}\right)^{3-6} &= \left(\dfrac{11}{9}\right)^{2x-1}\\[5pt] \left(\dfrac{11}{9}\right)^{-3} &= \left(\dfrac{11}{9}\right)^{2x-1}\\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ -3 &= 2x - 1 \\[5pt] -3 + 1 &= 2x \\[5pt] -2 &= 2x \\[5pt] \frac{-2}{2} &= x \\[5pt] x &= -1 \end{aligned} \]

Answer \( x = \color{red}{-1} \)

Solution

\[ \begin{aligned} 2^x \div 2^{-4} &= 4^5 \\[5pt] 2^{x - (-4)} &= (2^2)^5 \\[5pt] 2^{x + 4} &= 2^{10} \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ x + 4 &= 10 \\[5pt] x &= 10 - 4 \\[5pt] x &= 6 \end{aligned} \]

Answer \( x = \color{red} 6 \)

Solution

\[ \begin{aligned} &= (512)^{\frac{-2}{9}} \\[5pt] &= (2)^{\cancel9^1 \times \frac{-2}{\cancel9_1}} \\[5pt] &= 2^{-2} \\[5pt] &= \frac{1}{2^2} \\[5pt] &= \frac{1}{4} \end{aligned} \]

Answer \( \color{red} \dfrac{1}{4} \)

Solution

\[ \begin{align*} & = 4 \times 81^{\frac{-1}{2}} \, \bigl(81^{\frac{1}{2}}+81^{\frac{3}{2}}\bigr) \\ \\ & = 4 \times (9^2)^{\frac{-1}{2}} \, \Bigl[(9^2)^{\frac{1}{2}}+(9^2)^{\frac{3}{2}}\Bigr] \\ \\ & = 4 \times (9)^{\cancel2^1 \times \frac{-1}{\cancel2_1}} \, \Bigl[(9)^{\cancel2^1 \times \frac{1}{\cancel2_1}}+(9)^{\cancel2^1 \times \frac{3}{\cancel2_1}}\Bigr] \\ \\ & = 4 \times 9^{-1} \, \Bigl[9 ^1 + 9^3 \Bigr] \\ \\ & = 4 \times \Bigl[9^{-1} \times 9^1 + 9^{-1} \times 9^3 \Bigr] \\ \\ & = 4 \times \Bigl[9^{-1 + 1} + 9^{-1 + 3}\Bigr] \\ \\ & = 4 \times \Bigl[9^0 + 9^2\Bigr] \\ \\ & = 4 \times \Bigl[1 + 81 \Bigr] \\ \\ & = 4 \times 82 \\ \\ & = 328 \end{align*} \]

Answer \( 4\times81^{-\frac{1}{2}}(81^{\frac{1}{2}}+81^{\frac{3}{2}}) = \color{red}328\)

Solution

\[ \begin{align*} & = \frac{36^{\frac{7}{2}}-36^{\frac{9}{2}}}{36^{\frac{5}{2}}} \\ \\ & = \frac{(6^2)^{\frac{7}{2}}-(6^2)^{\frac{9}{2}}}{(6^2)^{\frac{5}{2}}} \\ \\ & = \frac{(6)^{\cancel2^1 \times \frac{7}{\cancel2_1}}-(6)^{\cancel2^1 \times \frac{9}{\cancel2_1}}}{(6)^{\cancel2^1 \times \frac{5}{\cancel2_1}}} \\ \\ & = \frac{6^7 -6^9}{6^5} \\ \\ & = \frac{6^7\,(1 -6^2)}{6^5} \\ \\ & = {6^{7-5}\,(1 - 36)} \\ \\ & = {6^{2}\,(-35)} \\ \\ & = 36\,(-35) \\ \\ & = -1260 \end{align*} \]

Answer \( \dfrac{36^{\frac{7}{2}}-36^{\frac{9}{2}}}{36^{\frac{5}{2}}} = \color{red}-1260\)

Solution

\[ \begin{align*} & = (0.04)^{\frac{3}{2}} \\ & = \Bigg(\frac{4}{100}\Bigg)^{\tfrac{3}{2}} \\ & = \Bigg(\frac{2^2}{10^2}\Bigg)^{\tfrac{3}{2}} \\ & = \Bigg[\Bigg(\frac{2}{10}\Bigg)^2 \ \Bigg]^{\tfrac{3}{2}} \\ & = \Bigg(\frac{2}{10}\Bigg)^{\cancel2^1 \times \tfrac{3}{\cancel2_1}} \\ & = \Bigg(\frac{2}{10}\Bigg)^3 \\ & = \Bigg(\frac{8}{1000}\Bigg) \\ \\ & = 0.008 \end{align*} \]

Answer \( (0.04)^{\frac{3}{2}} = \color{red} 0.008 \)

Solution

\[ \begin{align*} & = (0.03125)^{\frac{-2}{5}} \\ \\ & = \Bigg(\frac{3125}{100000}\Bigg)^{\tfrac{-2}{5}} \\ \\ & = \Bigg(\frac{100000}{3125}\Bigg)^{\tfrac{2}{5}} \\ \\ & = \Bigg[\Bigg(\frac{10^5}{5^5}\Bigg)\Bigg]^{\tfrac{2}{5}} \\ \\ & = \Bigg[\Bigg(\frac{10}{5}\Bigg)^5 \, \Bigg]^{\tfrac{2}{5}} \\ \\ & = \Bigg(\frac{10}{5}\Bigg)^{\cancel5^1 \times \tfrac{2}{\cancel5_1}} \\ \\ & = \Bigg(\frac{\cancel{10}^2}{\cancel5_1}\Bigg)^2 \\ \\ & = 2^2 \\ \\ & = 4 \end{align*} \]

Answer \( (0.03125)^{-\frac{2}{5}} = \color{red}4 \)

Solution

\[ \begin{align*} & = (1^3 + 2^3 + 3^3)^{\frac{1}{2}} \\ \\ & = (1 + 8 + 27)^{\frac{1}{2}} \\ \\ & = (36)^{\frac{1}{2}} \\ \\ & = (6^2)^{\frac{1}{2}} \\ \\ & = 6^{\cancel2^1 \times \frac{1}{\cancel2_1}} \\ \\ & = 6 \end{align*} \]

Answer \( (1^3 + 2^3 + 3^3)^{\frac{1}{2}} = \color{red} 6 \)

Solution

\[ \begin{align*} 4^x - 4^{x-1} & = 24 \\ \\ 4^x - \frac{4^x}{4} & = 24 \\ \\ 4^x \left( 1 - \frac{1}{4} \right) & = 24 \\ \\ 4^x \left( 1 - \frac{1}{4} \right) & = 24 \\ \\ 4^x \left(\frac{4 - 1}{4} \right) & = 24 \\ \\ 4^x \left(\frac{3}{4} \right) & = 24 \\ \\ 4^x & = \cancel{24}^8 \times \frac{4}{\cancel3_1} \\ \\ 4^x & = 32 \\[6pt] (2^2)^x & = 2^5 \\[6pt] 2^{2x} & = 2^5 \\[6pt] \text{Bases are the same, } & \text{we equate powers.} \\ 2x & = 5 \\[6pt] x & = \frac{5}{2} \\[6pt] \end{align*} \]

Answer \(x = \color{red} \dfrac{5}{2} \)

Solution

\[ \begin{aligned} \text{Let the number be } & x \\[5pt] 5^{-1} \times x &= (-7)^{-1} \\[8pt] \frac{1}{5} \times x &= \frac{-1}{7} \\[8pt] x &= \frac{-5}{7} \end{aligned} \]

Answer \( x = \color{red}{\dfrac{-5}{7}} \)

Solution

\[ \begin{aligned} \text{Let the number be } & x \\[5pt] (-15)^{-1} \div x &= (-5)^{-1} \\[8pt] (-15)^{-1} &= (-5)^{-1} \times x \\[8pt] (-15)^{-1} \div (-5)^{-1} &= x \\[8pt] \frac{-1}{15} \div \frac{-1}{5} &= x \\[8pt] \frac{-1}{15} \times (-5) &= x \\[8pt] \frac{\cancel5^1}{\cancel{15}_3} &= x \\[8pt] \frac{1}{3} &= x \\[8pt] \end{aligned} \]

Answer \( x = \color{red}{\dfrac{1}{3}} \)

Solution

\[ \begin{aligned} x &= \left(\frac{3}{2}\right)^2 \times \left(\frac{2}{3}\right)^{-4} \\[5pt] &= \left(\frac{3}{2}\right)^2 \times \left(\frac{3}{2}\right)^4 \\[5pt] &= \left(\frac{3}{2}\right)^{6} \\[5pt] x^{-2} &= \left[\left(\frac{3}{2}\right)^{6}\right]^{-2} \\[5pt] &= \left(\frac{3}{2}\right)^{-12} \\[5pt] &= \left(\frac{2}{3}\right)^{12} \end{aligned} \]

Answer \(x^{-2} = \color{red}{\left(\dfrac{2}{3}\right)^{12}}\)

Solution

\[ \begin{aligned} (81)^{-4}\div(729)^{2-x} &= 9^{4x} \\[5pt] (3^4)^{-4}\div(3^6)^{2-x} &= (3^2)^{4x} \\[5pt] 3^{-16}\div3^{6(2-x)} &= 3^{8x} \\[5pt] 3^{-16}\div3^{12-6x} &= 3^{8x} \\[5pt] 3^{-16-(12-6x)} &= 3^{8x} \\[5pt] 3^{-16-12+6x} &= 3^{8x} \\[5pt] 3^{6x-28} &= 3^{8x} \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ 6x-28 &= 8x \\[5pt] 28 &= 8x -6x \\[5pt] -28 &= 2x \\[5pt] \frac{-28}{2} &= x \\[5pt] x &= -14 \end{aligned} \]

Answer \(x = \color{red}{-14}\)

Solution

\[ \begin{aligned} \left(\frac{2}{3}\right)^x \times \left(\frac{3}{2}\right)^{2x} &= \frac{81}{16} \\[5pt] \left(\frac{3}{2}\right)^{-x} \times \left(\frac{3}{2}\right)^{2x} &= \frac{3^4}{2^4} \\[5pt] \left(\frac{3}{2}\right)^{-x + 2x} &= \left(\frac{3}{2}\right)^{4} \\[5pt] \left(\frac{3}{2}\right)^{x} &= \left(\frac{3}{2}\right)^{4} \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ x &= 4 \end{aligned} \]

Answer \(x = \color{red}{4}\)

Solution

\[ \begin{aligned} 6^x - 6^{x-2} &= 210 \\[5pt] 6^x - \frac{6^{x}}{6^2} &= 210 \\[5pt] 6^x \left(1 - \frac{1}{6^2} \right) &= 210 \\[5pt] 6^x \left(1 - \frac{1}{36} \right) &= 210 \\[5pt] 6^x \left(\frac{36- 1}{36} \right) &= 210 \\[5pt] 6^x \left(\frac{35}{36} \right) &= 210 \\[5pt] 6^x &= \cancel{210}^6 \times \frac{36}{\cancel{35}_1} \\[5pt] 6^x &= 6 \times 36 \\[5pt] 6^x &= 6 \times 6^2 \\[5pt] 6^x &= 6^3 \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ x &= 3 \end{aligned} \]

Answer \(x = \color{red}{3}\)

Solution

\[ \begin{aligned} & =\frac{(2^3\times3^4)^3\times(-5)^3}{60\times(-2)^5} \\[8pt] & =\frac{\cancel- (2^9 \times3^{12} \times 5^3)}{\cancel-(60\times 2^5)} \\[8pt] &= \frac{2^9 \times 3^{12} \times 5^3}{60\times 2^5} \\[8pt] &= \frac{2^9 \times 3^{12} \times 5^3}{2^2 \times 3 \times 5 \times 2^5} \\[8pt] &= \frac{2^9 \times 3^{12} \times 5^3}{2^7 \times 3 \times 5} \\[8pt] &= 2^{9-7} \times 3^{12-1} \times 5^{3-1} \\[8pt] &= 5^2\times2^2\times3^{11} \\[8pt] &= 25\times4\times177147 \\[8pt] &= 17714700 \end{aligned} \]

Answer \( \color{red}{17714700} \)

Solution

\[ \begin{aligned} & =\frac{3^5\times5^7\times(-2)^3}{3^4\times5^2\times(-2)^3} \\[5pt] &= 3^{5-4}\times5^{7-2}\times(-2)^{3-3} \\[5pt] &= 3^1\times5^5\times(-2)^0 \\[5pt] &= 3\times3125\times1 \\[5pt] &= 9375 \end{aligned} \]

Answer \( \color{red}{9375} \)