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Logarithmic Differentiation

Solve the fol...

Question

Solve the following equation.
${log}_{2} (4^{x} + 4) = {log}_{2} 2^{x} + {log}_{2} (2^{x + 1} - 3)$

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Solution

${log}_{2} (4^{x} + 4) = {log}_{2} 2^{x} + {log}_{2} (2^{x + 1} - 3) \Rightarrow (4^{x} + 4) = 2^{x} (2^{x + 1} - 3) \Rightarrow 4^{x} - {3.2}^{x} - 4 = 0$

But $2^{x}$ cannot be $< 0.$
Therefore, $2^{x} = 4 \Rightarrow x = 2.$

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