Solve the following equation:
$9^{{log}_{1 / 3} (x + 1)} = 5^{{log}_{1 / 5} (2 x^{2} + 1)}$

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Solution

$9 {log}_{\frac{1}{3}} (x + 1) = 5 {log}_{\frac{1}{5}} {(2 x}^{2} + 1) \Rightarrow {(x + 1)}^{{l o g}_{\frac{1}{3}}} 9 = {(2 x}^{2} + 1) {l o g}_{\frac{1}{5}} 5 \Rightarrow {(x + 1)}^{- 2} = {{(2 x}^{2} + 1)}^{- 1} \Rightarrow {(2 x}^{2} + 1) = {(x + 1)}^{2} \Rightarrow {2 x}^{2} + 1 = x^{2} + 2 x + 1 \Rightarrow x^{2} - 2 x = 0 \Rightarrow x = 0, 2.$

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