$\text{a)}$  

$5^x=6^{x+1}$  

$5^x=6^x\cdot 6\mid :6^x$  

$\frac{5^x}{6^x}=6$  

${\left(\frac{5}{6}\right)}^x=6$  

$x={\log }_{\frac{5}{6}}6$      

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$\text{b)}$  

${11}^{x-3}={13}^{3-x}$  

${11}^x\cdot {11}^{-3}={13}^3\cdot {13}^{-x}$  

${11}^x\cdot {\left(\frac{1}{11}\right)}^3={13}^3\cdot {\left(\frac{1}{13}\right)}^x\mid :{\left(\frac{1}{13}\right)}^x$  

${11}^x:{\left(\frac{1}{13}\right)}^x\cdot {\left(\frac{1}{11}\right)}^3={13}^3\mid :{\left(\frac{1}{11}\right)}^3$  

${11}^x:{\left(\frac{1}{13}\right)}^x={13}^3:{\left(\frac{1}{11}\right)}^3$  

${\left(11:\frac{1}{13}\right)}^x={\left(13:\frac{1}{11}\right)}^3$  

${\left(11\cdot 13\right)}^x={\left(13\cdot 11\right)}^3$  

${143}^x={143}^3$  

$x=3$  

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$\text{c)}$  

$7^{x+1}=3^{2x-2}$  

$7^x\cdot 7^1=3^{2x}\cdot 3^{-2}$  

$7^x\cdot 7={\left(3^2\right)}^x\cdot \frac{1}{9}$  

$7^x\cdot 7=9^x\cdot \frac{1}{9}\mid :9^x$  

$\frac{7^x}{9^x}\cdot 7=\frac{1}{9}\mid :7$  

${\left(\frac{7}{9}\right)}^x=\frac{1}{9}\cdot \frac{1}{7}$  

${\left(\frac{7}{9}\right)}^x=\frac{1}{63}$  

${\left(\frac{9}{7}\right)}^x=63$  

$x={\log }_{\frac{9}{7}}63$  

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$\text{d)}$  

$3^{2x}=2^x\cdot 4^{x-1}$  

${\left(3^2\right)}^x=2^x\cdot 4^x\cdot 4^{-1}$  

$9^x=2^x\cdot 4^x\cdot \frac{1}{4}$  

$9^x={\left(2\cdot 4\right)}^x\cdot \frac{1}{4}$  

$9^x=8^x\cdot \frac{1}{4}\mid :8^x$  

$\frac{9^x}{8^x}=\frac{1}{4}$  

${\left(\frac{9}{8}\right)}^x=\frac{1}{4}$  

$x={\log }_{\frac{9}{8}}\frac{1}{4}$  

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$\text{e)}$  

$4\cdot 2^{x-3}={10}^x\cdot 3^{x-1}$  

$4\cdot 2^x\cdot 2^{-3}={10}^x\cdot 3^x\cdot 3^{-1}$  

$4\cdot 2^x\cdot \frac{1}{8}={10}^x\cdot 3^x\cdot \frac{1}{3}$  

$\frac{1}{2}\cdot 2^x={\left(10\cdot 3\right)}^x\cdot \frac{1}{3}$  

$\frac{1}{2}\cdot 2^x={30}^x\cdot \frac{1}{3}\mid :{30}^x$  

$\frac{2^x}{{30}^x}\cdot \frac{1}{2}=\frac{1}{3}\mid :\frac{1}{2}$  

${\left(\frac{2}{30}\right)}^x=\frac{1}{3}\cdot \frac{2}{1}$  

${\left(\frac{1}{15}\right)}^x=\frac{2}{3}$  

${15}^x=\frac{3}{2}$  

$x={\log }_{15}\frac{3}{2}$  

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$\text{f)}$  

$5^{x+2}=16\cdot 2^{-x}$  

$5^x\cdot 5^2=16\cdot {\left(2^{-1}\right)}^x$  

$5^x\cdot 25=16\cdot {\left(\frac{1}{2}\right)}^x\mid :{\left(\frac{1}{2}\right)}^x$  

$\frac{5^x}{{\left(\frac{1}{2}\right)}^x}\cdot 25=16\mid :25$  

$5^x\cdot 2^x=\frac{16}{25}$  

${\left(5\cdot 2\right)}^x=\frac{16}{25}$  

${10}^x=\frac{16}{25}$  

$x={\log }_{10}\frac{16}{25}$