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Last updated date: 28th Nov 2023
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Show that x=2 is the only root of the equation${9^{{{\log }_3}[{{\log }_2}x]}} = {\log _2}x - {({\log _2}x)^2} + 1$

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Hint: Here let’s ${\text{use the properties }}{{\text{a}}^{{{\log }_a}^n}} = n$, ${\text{lo}}{{\text{g}}_a}x = b \Rightarrow x = {a^b},$ ${{\text{a}}^b}^{{{\log }_a}^n} = {a^{{{\log }_a}{n^b}}}$ and arrange the terms to find the value of x.

${\log _3}({\log _2}x){\text{ is defined only when lo}}{{\text{g}}_2}x = t(assumed){\text{ is + ve,i}}{\text{.e}}{\text{. , }}{\log _2}x > 0 = {2^1} \\ \therefore x > 1 \\ {\text{Also using the property }}{{\text{a}}^{{{\log }_a}^n}} = n \\ \Rightarrow {9^{{{\log }_3}t}} = {3^{2{{\log }_3}(t)}} = {3^{{{\log }_3}({t^2})}} = {t^2} \\ \therefore {t^2} = t - {t^2} + 1 \\$
${\text{ 2}}{{\text{t}}^2} - t - 1 = 0{\text{ }} \\$
${\text{(2t + 1)(t - 1) = 0}} \\ \therefore t = 1{\text{ only (}}\dfrac{{ - 1}}{2}{\text{ rejected as it is + ve)}} \\ \therefore {\text{lo}}{{\text{g}}_2}x = 1 \\ {\text{using the property lo}}{{\text{g}}_a}x = b \Rightarrow x = {a^b},{\text{ we get}} \\ {\text{x = }}{{\text{2}}^1} = 2 \\$
Thus there is only one root $2$.