Find the sum of all values of $x > 0$ for which ${({log}_{27} x^{3})}^{2} = {log}_{27} x^{6}$

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Solution

$(l o g_{27} x^{3})^{2} = (l o g_{27} x^{6})$
$\Rightarrow (3 l o g_{27} x^{2}) = 6 (l o g_{27} x) [∵ l o g a^{b} = b l o g a]$
$\Rightarrow 9 l o g_{27}^{x} [3 l o g_{27} x - 2] = 0$
$\Rightarrow l o g_{27} x = 0 \Rightarrow x = 1$
$l o g_{27} x = \frac{2}{3} \Rightarrow x = (27)^{2 / 3} = 9$
$∴ S u m o f v a l u e i s (9 + 1) = 110$

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