Solution

\begin{align*} &= 4^4 \times 5^{-4} \\ \\ &= 4^4 \times \left(\frac{1}{5}\right)^{4} \\ \\ &= \left(4 \times \frac{1}{5}\right)^{4} \\ \\ &= \left(\frac{4}{5}\right)^{4} \\ \\ &= \color{red} \frac{256}{625} \end{align*}

Solution

\begin{align*} &= 2^2 \times \left(-\frac{1}{3}\right)^2 \\ \\ &= \left(2 \times -\frac{1}{3}\right)^2 \\ \\ &= \left(-\frac{2}{3}\right)^2 \\ \\ &= \left(-\frac{2}{3}\right) \times \left(-\frac{2}{3}\right) \\ \\ &= \color{red} \frac{4}{9} \end{align*}

Solution

\begin{align*} &= \left(-\frac{2}{3}\right)^3 \times \left(-\frac{3}{5}\right)^3 \\ \\ &= \left(-\frac{2}{3} \times -\frac{3}{5}\right)^3 \\ \\ &= \left(\frac{2}{\cancel3} \times \frac{\cancel3}{5}\right)^3 \\ \\ &= \left(\frac{2}{5}\right)^3 \\ \\ &= \color{red} \frac{8}{125} \end{align*}

Solution

\begin{align*} & = \left(\frac{1}{2}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2} \\ \\ &= \left(\frac{2}{1}\right)^2 \times \left(\frac{5}{2}\right)^2 \\ \\ &= \left(\cancel2 \times \frac{5}{\cancel2}\right)^2 \\ \\ &= 5^2 \\ &= \color{red} 25 \end{align*}

Solution

\begin{align*} &= \left(-\frac{5}{6}\right)^4 \div \left(-\frac{7}{6}\right)^4 \\ \\ &= \left(-\frac{5}{6} \div -\frac{7}{6}\right)^4 \\ \\ &= \left(\frac{5}{\cancel6} \times \frac{\cancel6}{7}\right)^4 \\ \\ &= \left(\frac{5}{7}\right)^4 \\ \\ &= \color{red} \frac{625}{2401} \\ \\ \end{align*}

Solution

\begin{align*} & = \left(-\frac{2}{3}\right)^{-5} \times \left(-\frac{3}{2}\right)^{-5} \\ \\ &= \left(-\frac{3}{2}\right)^{5} \times \left(-\frac{2}{3}\right)^{5} \\ \\ &= \left(-\frac{3}{2} \times -\frac{2}{3}\right)^{5} \\ \\ &= \left(\frac{\cancel3}{\cancel2} \times \frac{\cancel2}{\cancel3}\right)^{5} \\ \\ &= 1^{5} \\ &= \color{red} 1 \end{align*}

Solution

\begin{align*} 3^2 \times (-4)^2 &= (-12)^{2x} \\ (3 \times -4)^2 &= (-12)^{2x} \\ (-12)^2 &= (-12)^{2x} \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 2x &= 2 \\ x &= \frac{\cancel2^1}{\cancel2_1} \\ \color{red} x &= \color{red}1 \end{align*}

Solution

\begin{align*} \left(-\frac{3}{2}\right)^6 \times {\color{magenta} \left(\frac{4}{9}\right)^3} &= \left(\frac{1}{2}\right)^{3x} \\ \\ \left(-\frac{3}{2}\right)^6 \times {\color{magenta}\left(\frac{2^2}{3^2}\right)^3} &= \left(\frac{1}{2}\right)^{3x} \\ \\ \left(-\frac{3}{2}\right)^6 \times {\color{magenta} \left[\left(\frac{2}{3}\right)^2\right]^3} &= \left(\frac{1}{2}\right)^{3x} \\ \\ \left(-\frac{3}{2}\right)^6 \times {\color{magenta} \left(\frac{2}{3}\right)^6} &= \left(\frac{1}{2}\right)^{3x} \\ \\ \left(-\frac{\cancel3}{\cancel2} \times \frac{\cancel2}{\cancel3}\right)^6 &= \left(\frac{1}{2}\right)^{3x} \\ \\ (-1)^6 &= \left(\frac{1}{2}\right)^{3x} \\ \\ 1 & = \left(\frac{1}{2}\right)^{3x} \\ \\ \left(\frac{1}{2}\right)^{0} &= \left(\frac{1}{2}\right)^{3x} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 3x &= 0 \\ \color{red} x &= \color{red} 0 \end{align*}

Solution

\begin{align*} \displaystyle \left(\frac{4}{5}\right)^{-2} \div \left(-\frac{4}{5}\right)^{-2} &= 1^{3x} \\ \\ \left(\frac{5}{4}\right)^{2} \div \left(-\frac{5}{4}\right)^{2} &= 1^{3x} \\ \\ \left[\frac{5}{4} \div \left(-\frac{5}{4}\right)\right]^{2} &= 1^{3x} \\ \\ \left[\frac{\cancel5^1}{\cancel4_1} \times \left(-\frac{\cancel4^1}{\cancel5_1}\right)\right]^{2} &= 1^{3x} \\ \\ (-1)^{2} &= 1^{3x} \\ 1^1 &= 1^{3x} \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 3x & = 1 \\ \color{red} x & = \color{red} \frac{1}{3} \end{align*}

Solution

\begin{align*} \left(\frac{9}{4}\right)^3 \times \left(\frac{8}{9}\right)^3 &= 2^{6x} \\ \\ \left(\frac{\cancel9^1}{\cancel4_1} \times \frac{\cancel8^2}{\cancel9_1}\right)^3 &= 2^{6x} \\ \\ 2^3 &= 2^{6x} \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 6x &= 3 \\ x &= \frac{\cancel3^1}{\cancel6_2} \\ \color{red} x &= \color{red} \frac{1}{2} \end{align*}

Solution

\begin{align*} \left(\frac{15}{4}\right)^3 \div \left(\frac{5}{4}\right)^3 &= 3^x \\ \\ \left(\frac{15}{4} \div \frac{5}{4}\right)^3 &= 3^x \\ \\ \left(\frac{\cancel{15}^3}{\cancel4_1} \times \frac{\cancel4^1}{\cancel5_1}\right)^3 &= 3^x \\ 3^3 &= 3^x \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ \color{red} x &= \color{red} 3 \end{align*}