Question

Solve the following equation : ${3^{{{\left( {{{\log }_9}x} \right)}^2} - \dfrac{9}{2}{{\log }_9}x + 5}} = 3\sqrt 3$

Hint: We are going to convert the given equation into a quadratic equation by comparing the powers on both sides and then by applying the basic properties of logarithm the values of x are computed.

Given ${3^{{{\left( {{{\log }_9}x} \right)}^2} - \dfrac{9}{2}{{\log }_9}x + 5}} = 3\sqrt 3$
Comparing the powers on both sides of the given equation, we get
${\left( {{{\log }_9}x} \right)^2} - {\log _9}x + 5 = \dfrac{3}{2}$
Let $\left( {{{\log }_9}x} \right) = y$
$\therefore {y^2} - \dfrac{9}{2}y + 5 = \dfrac{3}{2}$
$\Rightarrow 2{y^2} - 9y + 10 - 3 = 0$
$\Rightarrow 2{y^2} - 9y + 7 = 0$
$\Rightarrow 2{y^2} - 7y - 2y + 7 = 0$
$\Rightarrow y(2y - 7) - (2y - 7) = 0$
$\Rightarrow (y - 1)(2y - 7) = 0$
$\therefore y = 1,\dfrac{7}{2}$
When $y = 1$
$\therefore {\log _9}x = 1$
$\Rightarrow x = 9{\text{ }}[\because {\log _b}x = y \Leftrightarrow x = {b^y}]$
When $y = \dfrac{7}{2}$
$\therefore {\log _9}x = \dfrac{7}{2}$
$\Rightarrow x = {\left( 9 \right)^{\dfrac{7}{2}}}{\text{ }}[\because {\log _b}x = y \Leftrightarrow x = {b^y}] \\ \Rightarrow x = {\left( {{3^2}} \right)^{\dfrac{7}{2}}} \\ \Rightarrow x = {(3)^7} \\$
$\therefore x = 9,{(3)^7}$

Note: To solve the given problems on logarithms. The basic properties and formulae should be known. Here, we used the basic property of logarithm ${\log _b}x = y \Leftrightarrow x = {b^y}$ i.e.., conversion of a logarithm form into exponential form.