Solve the following equation:
${log}_{3} (9^{x} + 9) = x + {log}_{3} (28 - {2.3}^{x})$

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Solution

${log}_{3} (9^{x} + 9) = x + {log}_{3} (28 - {2.3}^{x}) \Rightarrow 9^{x} + 9 = 3^{x} (28 - {2.3}^{x})$

Let $3^{x} = a$
$a^{2} + 9 = a (28 - 2 a) \Rightarrow a^{2} + 9 = 28 a - {2 a}^{2} \Rightarrow {3 a}^{2} - 28 a + 9 = 0$

Therefore, $a = 9, \frac{1}{3}$
$3^{x} = 9, \frac{1}{3}$
Therefore, $x = 2, - 1$

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