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

10

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7

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8

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9

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Solution

The correct option is B $10$
$l o g_{27} x^{3} = l o g_{3} x$ and $l o g_{27} x^{6} = 2 l o g_{3} x$
Now, let $l o g_{3} x = z$
$z^{2} = 2 z$
$(z^{2} - 2 z) = 0$
$z = 2, 0$
$z = 2 \Rightarrow l o g_{3} x = 2 \Rightarrow x = 9$
$z = 0 \Rightarrow l o g_{3} x = 0 \Rightarrow x = 1$
sum of values of $x = 10$

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