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সমাধান কর :

\(3^{x + 2} = 81\)
\(5^{3x - 7} = 3^{3x - 7}\)
\(2^{x - 4} = 4a^{x - 6}\;\;(a > 0,\, a \ne 2)\)
\((\sqrt{3})^{x + 5} = (\sqrt[3]{3})^{2x + 5}\)
\((\sqrt[5]{4})^{4x + 7} = (\sqrt[4]{64})^{2x + 7}\)
\(\dfrac{3^{3x - 4} \cdot a^{2x - 5}}{3^{x + 1}} = a^{2x - 5}\;\;(a > 0)\)
\(\dfrac{5^{2x} \cdot b^{x - 3}}{5^{x + 3}} = a^{x - 3}\;\;(a,\, b > 0,\; 5b \ne a)\)
\(4^{x + 2} = 2^{2x + 1} + 14\)
\(5^x + 5^{2 - x} = 26\)
\(3(9^x - 4 \cdot 3^{x - 1}) + 1 = 0\)
\(4^{1 + x} + 4^{1 - x} = 10\)
\(2^{2x} - 3 \cdot 2^{x + 2} = -32\)