Question

The equation $2{\log _2}({\log _2}x) + {\log _{1/2}}{\log _2}(2\sqrt {2x} ) = 1$ has
A. product of all its solution =4
B. a rational solution which is not an integer
C. has a natural solution
D. has no prime solutions

Answer 1

Here, we would be using the basic properties of logarithmic function. Then find the value of x from the quadratic equation.

Complete step-by-step answer:
Given, $2{\log _2}({\log _2}x) + {\log _{1/2}}{\log _2}(2\sqrt {2x} ) = 1$
(using the property which is given by $a\log b = \log {b^a}$and $\log _a^b = \dfrac{{\log b}}{{\log a}}$)
$\Rightarrow {\log _2}{({\log _2}x)^2} + \dfrac{{\log \left( {{{\log }_2}(2\sqrt {2x} )} \right)}}{{\log 1/2}} = 1$
$\Rightarrow \dfrac{{\log {{({{\log }_2}x)}^2}}}{{\log 2}} + \dfrac{{\log \left( {{{\log }_2}(2\sqrt {2x} )} \right)}}{{\log 1/2}} = 1$
$\Rightarrow \dfrac{{\log {{({{\log }_2}x)}^2}}}{{\log 2}} - \dfrac{{\log \left( {{{\log }_2}(2\sqrt {2x} )} \right)}}{{\log 2}} = 1$(using the property which is given by $\log \dfrac{a}{b} = - \log \dfrac{b}{a}$)
$\Rightarrow \dfrac{{\log {{({{\log }_2}x)}^2} - \log \left( {{{\log }_2}(2\sqrt {2x} )} \right)}}{{\log 2}} = 1$
$\Rightarrow \log \left( {\dfrac{{{{({{\log }_2}x)}^2}}}{{{{\log }_2}(2\sqrt {2x} )}}} \right) = \log 2$ (using the property which is given by $\log a - \log b = \log \dfrac{a}{b}$)
$\Rightarrow \log \left( {\dfrac{{{{({{\log }_2}x)}^2}}}{{{{\log }_2}({2^{3/2}}x)}}} \right) = \log 2$
$\Rightarrow \log \left( {\dfrac{{{{({{\log }_2}x)}^2}}}{{{{\log }_2}({2^{3/2}}) + {{\log }_2}x}}} \right) = \log 2$ (using the property which is given by $\log a + \log b = \log ab$)
$\Rightarrow \log {({\log _2}x)^2} = \log 2\left( {\dfrac{3}{2} + {{\log }_2}x} \right)$(using the property which is given by${\log _a}{a^n} = n$)

$$\Rightarrow {({\log _2}x)^2} = 2\left( {\dfrac{3}{2} + {{\log }_2}x} \right) \\\ \Rightarrow {({\log _2}x)^2} = 3 + 2{\log _2}x \\\ \Rightarrow {({\log _2}x)^2} - 2{\log _2}x - 3 = 0 \\\ \Rightarrow {({\log _2}x)^2} - 3{\log _2}x + {\log _2}x - 3 = 0 \\\ \Rightarrow {\log _2}x({\log _2}x - 3) + 1({\log _2}x - 3) = 0 \\\ \Rightarrow ({\log _2}x + 1)({\log _2}x - 3) = 0 \\\ \Rightarrow x = \dfrac{1}{2},x = {2^3} = 8 \\\$$

Now if we take the product of $x = \dfrac{1}{2}$and $x = 8$then we get 4
Therefore, option A. product of all its solution =4 is the required solution
Note: (i) The properties of logarithmic function should be used carefully.
(ii) One should avoid common mistakes such as $r{\log _a}M \ne {({\log _a}M)^r}$