Question

The product of all values of $x$ which makes the following statement true. $\left( {{{\log }_3}x} \right)\left( {{{\log }_5}9} \right) - \left( {{{\log }_x}25} \right) + \left( {{{\log }_3}2} \right) = \left( {{{\log }_3}54} \right)$ is

Answer 1

Hint: In this question, we are given a logarithmic equation which we have to solve for all possible values of $x$ and then find the product of those values.
So, we will use properties of logarithmic functions to simplify the equation and find the value of $x$ .
We will use properties of a logarithmic function which will give us a quadratic equation, in $t$ say, (for simplicity) and then solve that equation for $t$ .
Then, we will substitute the value of $t$ in terms of $x$ , and then find the value of $x$ .

Formulae to be used:
$\log (ab) = \log a + \log b$
$\log {a^n} = n\log a$
${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$
${\log _a}b = \dfrac{{\log b}}{{\log a}}$
If ${x^n} = A$ , then ${\log _x}A = n$ .

Complete answer:
We are given a logarithmic equation: $\left( {{{\log }_3}x} \right)\left( {{{\log }_5}9} \right) - \left( {{{\log }_x}25} \right) + \left( {{{\log }_3}2} \right) = \left( {{{\log }_3}54} \right)$ .
To find and solve the equation for $x$ .
So, first we’ll write the given equation i.e., $\left( {{{\log }_3}x} \right)\left( {{{\log }_5}9} \right) - \left( {{{\log }_x}25} \right) + \left( {{{\log }_3}2} \right) = \left( {{{\log }_3}54} \right)$ .
We can write the equation as
$\left( {{{\log }_3}x} \right)\left( {{{\log }_5}{3^2}} \right) - \left( {{{\log }_x}{5^2}} \right) + \left( {{{\log }_3}2} \right) = \left( {{{\log }_3}{3^3} \cdot 2} \right)$ .

Now, using the identity $\log (ab) = \log a + \log b$ ,
 we can write the equation as
$\left( {{{\log }_3}x} \right)\left( {{{\log }_5}{3^2}} \right) - \left( {{{\log }_x}{5^2}} \right) + \left( {{{\log }_3}2} \right) = \left( {{{\log }_3}{3^3}} \right) + {\log _3}2$ .

Now, using the identity $\log {a^n} = n\log a$ , we can write the equation as
$\left( {{{\log }_3}x} \right)\left( {2{{\log }_5}3} \right) - \left( {2{{\log }_x}5} \right) + \left( {{{\log }_3}2} \right) = \left( {3{{\log }_3}3} \right) + {\log _3}2$ .

Now, using the identity, ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ , we can write the above equation as
\[\dfrac{{\left( {\log x} \right)}}{{\left( {\log 3} \right)}} \times \dfrac{{2\log 3}}{{\log 5}} - \left( {2{{\log }_x}5} \right) + \left( {{{\log }_3}2} \right) = \left( {3{{\log }_3}3} \right) + {\log _3}2\] , canceling $\log 3$ by $\log 3$ gives,

$\dfrac{{2\log x}}{{\log 5}} - 2{\log _x}5 + {\log _3}2 = 3{\log _3}3 + {\log _3}2$ , and canceling ${\log _3}2$ by ${\log _3}2$ and we have ${\log _3}3 = 1$ , we get, $\dfrac{{2\log x}}{{\log 5}} - 2{\log _x}5 = 3{\log _3}3$ , i.e., $2{\log _5}x - 2{\log _x}5 = 3$ , which can be written as $2{\log _5}x - \dfrac{2}{{{{\log }_5}x}} - 3 = 0$ .

Now, let ${\log _5}x$ be $t$ , then the equation becomes, $2t - \dfrac{2}{t} - 3 = 0$ i.e., $2{t^2} - 3t - 4 = 0$ .

Now, solving the equation using the middle term split method gives, $t = - \dfrac{1}{2},2$ .
Substituting the value of $t$ , we get, ${\log _5}x = - \dfrac{1}{2},2$ .

Now, we know that if ${x^n} = A$ , then ${\log _x}A = n$ , so, we have that, $x = {5^{ - \dfrac{1}{2}}},{5^2}$ i.e., $x = \dfrac{1}{{\sqrt 5 }},25$ .

Hence,the product of the values $ = \dfrac{1}{{\sqrt 5 }} \times 25 = \dfrac{{5 \times 5}}{{\sqrt 5 }}$ i.e., $5\sqrt 5 $

Note: One must know all the properties of logarithmic function for solving such equations.
The quadratic equation can also be solved using the ‘Discriminant method’, in which, for a quadratic equation $a{x^2} + bx + c = 0$ , we have, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , on putting values of $a,b,c$ , we will get the same value for $x$ .
Canceling the same terms from each of the equality, is the same as subtracting the same terms from each side of the equality.