Solution

\begin{align*} \begin{array}{r|l} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} && \begin{array}{r|l} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \end{align*}\begin{aligned} 256 &= 2^8 \\ 64 &= 2^6\\ 729 &= 3^6 \\ 81 &= 3^4 \\ \\ &= \frac{256 \times 81}{64 \times 729} \\ \\ &= \frac{2^8 \times 3^4}{2^6 \times 3^6} \\ \\ &= \frac{2^{8-6}}{3^{6-4}} \\ \\ &= \frac{2^{2}}{3^{2}} \\ \\ &= \left(\frac{2}{3}\right)^2 \\ \\ \end{aligned}

Answer \( \displaystyle \frac{256 \times 81}{64 \times 729} =\color{red} \left(\frac{2}{3}\right)^2 \)

Solution

\[ \begin{aligned} 64 &= 2^6 \\ \\ 2^{2x-3} &= (2^6)^x \\ 2^{2x-3} &= 2^{6x} \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ 2x - 3 &= 6x \\ 2x - 6x &= 3 \\ -4x &= 3 \\ x &= \frac{-3}{4} \\ \end{aligned} \]

Answer Value of \( x = \displaystyle \color{red} \frac{-3}{4} \)