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Show that (2x-10)/(10x-1-5x) can be reduced to 10/5x

$\frac{2^x\;-\;10}{10^{x-1}\;-\;5^x} \\ =\frac{2^x\;-\;10}{10^{x}10^{-1}\;-\;5^x} \\ =\frac{2^x\;-\;10}{2^{x}*5^x*10^{-1}\;-\;5^x} \\ =\frac{2^x\;-\;10}{5^x(2^{x}*10^{-1}\;-\;1)} \\ =\frac{2^x\;-\;10}{5^x(\frac{2^{x}}{10}\;-\;\frac{10}{10})} \\ =\frac{2^x\;-\;10}{\frac{5^x}{10}(2^x\;-\;10)} \\ =\frac{1}{\frac{5^x}{10}} \\ =1\div{\frac{5^x}{10}} \\ ={\frac{10}{5^x}} \\$

Show that (2x-10)/(10x-1-5x) can be reduced to 10/5x

$\begin{array}{rcll} && \frac{2^x-10} {10^{x-1}-5^x } \qquad | \qquad \cdot \frac{2^x}{2^x}\\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x(10^{x-1}-5^x) } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- 2^x\cdot 5^x } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- (2\cdot 5)^x } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- 10^x } \\\\ &=& \frac{2^x\cdot (2^x-10)} {2^x\cdot 10^{x-1}- 10\cdot 10^{x-1} } \\\\ &=& \frac{2^x\cdot (2^x-10)} { 10^{x-1} ( 2^x - 10 ) } \\\\ &=& \frac{2^x} { 10^{x-1} } \\\\ &=& \frac{2^x} { \frac{10^x}{10} } \\\\ &=& 10\cdot \frac{2^x}{10^x} \\\\ &=& 10\cdot \left(\frac{2}{10} \right)^x \\\\ &=& 10\cdot \left(\frac{1}{5} \right)^x \\\\ &=& 10\cdot \left(\frac{1}{5^x} \right) \\\\ &=& \dfrac{10}{5^x} \end{array}$