\begin{align*} 5(2^x) + 2^{x + 1} - 3(2^x) & = 5(2^x) + 2^x \times 2^1 - 3(2^x) \phantom{000000} [a^{m + n} = a^m \times a^n] \\ & = 5(2^x) + 2(2^x) - 3(2^x) \phantom{00000000} [5y + 2y - 3y = 4y] \\ & = 4(2^x) \\ & = 2^2 \times 2^x \\ & = 2^{x + 2} \end{align*}

\begin{align*} \sqrt{(4x^{-2}y^6)^3} & = (4x^{-2} y^6)^{3 \over 2} \phantom{0000000000000} [ \sqrt[n]{a^m} = a^{m \over n}] \\ & = 4^{3 \over 2} (x^{-2})^{3 \over 2} (y^6)^{3 \over 2} \phantom{000000000}[ (ab)^m = a^m \times b^m] \\ & = 8 x^{-3} y^9 \phantom{0000000000000000} [ (a^m)^n = a^{mn}] \\ & = \left(8 \over 1\right) \left(1 \over x^3\right) \left(y^9 \over 1\right) \phantom{0000.} \left[ a^{-n} = {1 \over a^n} \right] \\ & = {8y^9 \over x^3} \end{align*}

\begin{align*} \left( {9x^{-4}y^6 \over 25x^2y^{-2}} \right)^{-{1 \over 2}} & = \left( { 25x^2 y^{-2} \over 9x^{-4} y^6 } \right)^{1 \over 2} \phantom{000000000000} \left[ \left(a \over b\right)^{-n} = \left(b \over a\right)^n \right] \\ & = { (25)^{1 \over 2} (x^2)^{1 \over 2} (y^{-2})^{1 \over 2} \over (9)^{1 \over 2} (x^{-4})^{1 \over 2} (y^6)^{1 \over 2} } \\ & = { 5 x^1 y^{-1} \over 3 x^{-2} y^3} \phantom{000000000000000000} [ (a^m)^n = a^{mn}] \\ & = {5 \over 3} x^{1 - (-2)} y^{-1 - 3} \phantom{00000000000.} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = {5 \over 3} x^3 y^{-4} \\ & = {5 \over 3} \left(x^3 \over 1\right) \left(1 \over y^4\right) \phantom{0000000000} \left[ a^{-n} = {1 \over a^n} \right] \\ & = {5x^3 \over 3y^4} \end{align*}

\begin{align*} {4^x + 4^x \over 2^{1 - 2x} } & = { 2(4^x) \over 2^{1 - 2x}} \phantom{000000000} [y + y = 2y \implies 4^x + 4^x = 2(4^x)] \\ & = { 2[ (2^2)^x ] \over 2^{1 - 2x} } \\ & = { 2^1 \times 2^{2x} \over 2^{1 - 2x} } \phantom{0000000} [(a^m)^n = a^{mn}] \\ & = { 2^{1 + 2x} \over 2^{1 - 2x} } \phantom{000000000.} [ a^m \times a^n = a^{m + n} ] \\ & = 2^{ 1 + 2x - (1 - 2x) } \phantom{000.0} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = 2^{ 1 + 2x - 1 + 2x } \\ & = 2^{4x} \end{align*}

\begin{align*} {4x^{-2} y^3 \over 3z^4} \div {2x^4 y^2 \over (3yz)^2 } & = {4x^{-2} y^3 \over 3z^4} \times {(3yz)^2 \over 2x^4 y^2} \\ & = {4x^{-2} y^3 \over 3z^4} \times { 3^2 y^2 z^2 \over 2x^4 y^2} \\ & = {4x^{-2} y^3 \over 3z^4} \times { 9 y^2 z^2 \over 2x^4 y^2} \\ & = {36 x^{-2} y^5 z^2 \over 6 x^4 y^2 z^4 } \\ & = {36 \over 6} \left(x^{-2} \over x^4\right) \left(y^5 \over y^2\right) \left(z^2 \over z^4\right) \\ & = 6 (x^{-6})(y^3)(z^{-2}) \\ & = {6 \over 1} \left(1 \over x^6\right)\left(y^3 \over 1\right) \left(1 \over z^2\right) \\ & = {6y^3 \over x^6 z^2} \end{align*}

\begin{align*} 3^{2x - 1} & = {1 \over 81} \\ 3^{2x - 1} & = 81^{-1} \phantom{000000000} \left[ {1 \over a^n} = a^{-n} \right] \\ 3^{2x - 1} & = (3^4)^{-1} \\ 3^{2x - 1} & = 3^{-4} \phantom{00000000000} [(a^m)^n = a^{mn}] \\ \\ 2x - 1 & = -4 \\ 2x & = -4 + 1 \\ 2x & = -3 \\ x & = -{3 \over 2} \end{align*}

\begin{align*} 5^a + 5^{a + 1} & = 750 \\ 5^a + 5^a \times 5^1 & = 750 \phantom{000000} [a^{m + n} = a^m \times a^n] \\ 5^a + 5(5^a) & = 750 \phantom{000000} [ x + 5x = 6x \implies 5^a + 5(5^a) = 6(5^a)] \\ 6(5^a) & = 750 \\ 5^a & = {750 \over 6} \\ 5^a & = 125 \\ 5^a & = 5^3 \\ \\ a & = 3 \end{align*}