Find the sum of all the values of $x$ satisfying ${log}_{x^{2} - 1} (x^{3} + 6) = {log}_{x^{2} - 1} (4 x^{2} - 4)$

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Solution

${log}_{x^{2} - 1} (x^{3} + 6) = {log}_{x^{2} - 1} (4 x^{2} - 4)$

$\Rightarrow {log}_{x^{2} - 1} (x^{3} + 6) - {log}_{x^{2} - 1} (4 x^{2} - 4) = 0$

$\Rightarrow {log}_{x^{2} - 1} {\frac{x^{3} + 6}{4 x^{2} - 4}} = 0$

$\Rightarrow \frac{x^{3} + 6}{4 (x^{2} - 1)} = 1$

$\Rightarrow x^{3} + 6 = 4 x^{2} - 4$

$\Rightarrow x^{3} - 4 x^{2} + 10 = 0$

$\Rightarrow$ Sum of all the roots $= \frac{- (b)}{a} = \frac{- (- 4)}{1} = 4$

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