$\text{a)}{2}^{17}\cdot x-{16}^4\cdot 3=5\cdot \left(4^8\cdot x-3\cdot 2^{17}\right)$  

$2^{17}\cdot x-{\left(2^4\right)}^4\cdot 3=5\cdot \left[{\left(2^2\right)}^8\cdot x-3\cdot 2^{17}\right]$  

$2^{17}\cdot x-2^{16}\cdot 3=5\cdot \left(2^{16}\cdot x-3\cdot 2^{17}\right)$  

$2^{16}\cdot 2\cdot x-2^{16}\cdot 3=5\cdot \left(2^{16}\cdot x-3\cdot 2\cdot 2^{16}\right)$  

$2^{16}\cdot 2x-2^{16}\cdot 3=5\cdot \left(2^{16}\cdot x-6\cdot 2^{16}\right)$  

$2^{16}\cdot \left(2x-3\right)=5\cdot 2^{16}\cdot \left(x-6\right)\mid :2^{16}$  

$2x-3=5\left(x-6\right)$  

$2x-3=5x-30$  

$-3x=-27\mid :\left(-3\right)$  

$x=9$  

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$\text{b)}{27}^3\cdot 2x-3^9=3^{10}\cdot x+2\cdot 3^9$  

${\left(3^3\right)}^3\cdot 2x-3^9=3^9\cdot 3\cdot x+2\cdot 3^9$  

$3^9\cdot 2x-1\cdot 3^9=3^9\cdot 3x+2\cdot 3^9$  

$3^9\cdot \left(2x-1\right)=3^9\cdot \left(3x+2\right)\mid :3^9$  

$2x-1=3x+2$  

$-x=3\mid \cdot \left(-1\right)$  

$x=-3$  

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$\text{c)}{3}^{13}\cdot x-4\cdot 9^6={27}^4\cdot \left(2x-5\right)$  

$3^{13}\cdot x-4\cdot {\left(3^2\right)}^6={\left(3^3\right)}^4\cdot \left(2x-5\right)$  

$3^{12}\cdot 3\cdot x-4\cdot 3^{12}=3^{12}\cdot \left(2x-5\right)$  

$3^{12}\cdot \left(3x-4\right)=3^{12}\cdot \left(2x-5\right)\mid :3^{12}$  

$3x-4=2x-5$  

$x=-1$  

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$\text{d)}\left(4^5\cdot x+{32}^2\right)\cdot 3^5=2^{16}\cdot x$  

$\left[{\left(2^2\right)}^5\cdot x+{\left(2^5\right)}^2\right]\cdot 2^5=2^{16}\cdot x\mid :2^5$  

$2^{10}\cdot x+2^{10}=2^{11}\cdot x$  

$2^{10}\cdot x+1\cdot 2^{10}=2^{10}\cdot 2x$  

$2^{10}\left(x+1\right)=2^{10}\cdot 2x\mid :2^{10}$  

$x+1=2x$  

$x=1$  

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$\text{e)}\frac{x}{2^5}+{\left(\frac{1}{4}\right)}^2={\left(-\frac{1}{8}\right)}^2\cdot x+\frac{1}{2^3}$  

$\frac{x}{2^5}+\frac{1}{{\left(2^2\right)}^2}=\frac{1}{{\left(2^3\right)}^2}\cdot x+\frac{1}{2^3}$  

$\frac{x}{2^5}+\frac{1}{2^4}=\frac{x}{2^6}+\frac{1}{2^3}\mid \cdot 2^6$  

$2x+2^2=x+2^3$  

$2x+4=x+8$  

$x=4$  

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$\text{f)}\frac{x}{4^4}-\frac{1}{{\left(-2\right)}^6}=-{\left(-\frac{1}{16}\right)}^2$  

$\frac{x}{4^4}-\frac{1}{2^6}=-{\left(\frac{1}{16}\right)}^2$  

$\frac{x}{{\left(2^2\right)}^4}-\frac{1}{2^6}=-\frac{1}{{\left(2^4\right)}^2}$  

$\frac{x}{2^8}-\frac{1}{2^6}=-\frac{1}{2^8}\mid \cdot 2^8$  

$x-2^2=-1$  

$x-4=-1$  

$x=3$