Question

Show that x=2 is the only root of the equation
${9^{{{\log }_3}[{{\log }_2}x]}} = {\log _2}x - {({\log _2}x)^2} + 1$

Answer 1

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.

Complete step-by-step answer:
Here we have
${\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 \\\$
Now re - arranging the terms, we get
${\text{ 2}}{{\text{t}}^2} - t - 1 = 0{\text{ }} \\\$
Splitting the middle terms, we get
${\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 $2$.

Note: The properties used above are very important for other problems as well, and many more properties of logarithm functions are present. One must remember all the properties to know the approach towards the solution.