If ${log}_{3} [l o g_{2} (l o g_{3} x)] = 1$ , show that $x = 6561$ .

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Solution

We know that,

${log}_{a} b = x \Rightarrow b = a^{x}$

$l o g_{3} [l o g_{2} (l o g_{3} x)] = 1$

$l o g_{2} (l o g_{3} x) = 3^{1} = 3$

$l o g_{3} x = 2^{3} = 8$

$x = 3^{8} = 6561$ .

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