For how many integral values of a does the equation ${log}_{e} (x^{2} + 2 a x) = {log}_{e} (8 x - 6 a - 3)$ have only one solution ?

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Solution

${log}_{e} (x^{2} + 2 a x) = {log}_{e} (8 x - 6 a - 3)$
${log}_{e} (x^{2} + 2 a x) - {log}_{e} (8 x - 6 a - 3) = 0$
$\Rightarrow {log}_{e} (\frac{x^{2} + 2 a x}{8 x - 6 a - 3}) = 0$
$\Rightarrow \frac{x^{2} + 2 a x}{8 x - 6 a - 3} = 1$
$\Rightarrow x^{2} + (2 a - 8) x + 6 a + 3 = 0$
For exactly one solution,
$D = 0$
$\Rightarrow (2 a - 8)^{2} = 4 (6 a + 3)$
$\Rightarrow a^{2} - 14 a + 13 = 0$
$\Rightarrow a = 1, 13$
So, 2 integral values

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l o g (x^{2} + 2 a x) = l o g (8 x - 6 a - 3)

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