Solve the following equation for x:
$3^{2 x + 4} + 1 = {2.3}^{x + 2}$

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Solution

Given,

$3^{2 x + 4} + 1 = 2 \times 3^{x + 2}$

$3^{2 x} \times 3^{4} + 1 = 2 \times 3^{x} \times 3^{2}$

${(3^{x})}^{2} \times 3^{4} + 1 = 2 \times 3^{x} \times 3^{2}$

substitute $u = 3^{x}$

${(u)}^{2} \times 3^{4} + 1 = 2 u \times 3^{2}$

$81 u^{2} - 18 u + 1 = 0$

solving the above equation, we get,

$u = \frac{1}{9}$

$3^{x} = \frac{1}{9}$

$∴ x = - 2$

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Similar questions

Solve the following equations for x:

(i) $2^{2 x} - 2^{x + 3} + 2^{4}$ =0

(ii) $3^{2 x + 4} + 1 = 2 . 3^{x + 2}$

\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5 x}; x \neq 0, - 1, 2

3^{2 x + 4} + 1 = {2.3}^{x + 2}

5^{2 x} = 3^{2 x} + {2.5}^{x} + {2.3}^{x}

x

x + 3 (2 x + 5) = - 20

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