Solve the following equation:
$lo g_{2} (4^{x} + 1) = x + lo g_{2} (2^{x + 3} - 6)$

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Solve the following equation:

lo g_{2} (4^{x} + 1) = x + lo g_{2} (2^{x + 3} - 6)

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Text solutionVerified

lo g_{2} (4^{x} + 1) = x + lo g_{2} (2^{x + 3} - 6) \Rightarrow lo g_{2} (4^{x} + 1) = lo g_{2} 2^{x} + lo g_{2} (2^{x + 3} - 6) \Rightarrow 4^{x} + 1 = 2^{x} (2^{x + 3} - 6) \Rightarrow 4^{x} + 1 = 8.4^{x} - 6.2^{x} \Rightarrow 7. 4^{x} - 6.2^{x} - 1 = 0 \Rightarrow 2^{x} = 1, - \frac{1}{7}

But

2^{x}

never be negative.

Therefore

2^{x} = 1 \Rightarrow x = 0

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Question Text	Solve the following equation: $lo g_{2} (4^{x} + 1) = x + lo g_{2} (2^{x + 3} - 6)$
Answer Type	Text solution:1
Upvotes	150