The solution of the equation 4 cdot 9^{x - 1} | vedclass

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Question

(C) Given equation: $4 \cdot 9^{x - 1} = 3 \cdot \sqrt{2^{2x + 1}}$Rewrite the equation: $2^2 \cdot (3^2)^{x - 1} = 3^1 \cdot (2^{2x + 1})^{1/2}$$2^2

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(C) Given equation: $4 \cdot 9^{x - 1} = 3 \cdot \sqrt{2^{2x + 1}}$Rewrite the equation: $2^2 \cdot (3^2)^{x - 1} = 3^1 \cdot (2^{2x + 1})^{1/2}$$2^2 \cdot 3^{2x - 2} = 3^1 \cdot 2^{x + 0.5}$Divide both sides by $3^1$ and $2^{x + 0.5}$:$\frac{2^2}{2^{x + 0.5}} = \frac{3^1}{3^{2x - 2}}$$2^{2 - (x + 0.5)} = 3^{1 - (2x - 2)}$$2^{1.5 - x} = 3^{3 - 2x}$$2^{1.5 - x} = 3^{2(1.5 - x)}$$2^{1.5 - x} = (3^2)^{1.5 - x}$$2^{1.5 - x} = 9^{1.5 - x}$This equation holds true only if the exponent is zero:$1.5 - x = 0$$x = 1.5$

The solution of the equation $4 \cdot 9^{x - 1} = 3 \cdot \sqrt{2^{2x + 1}}$ is

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