# The number of value(s) of x satisfying the equation ${4^{{{\log }_2}\log x}} - 1 + {\ln ^3}x - 3{\ln ^2}x - 5\ln x + 7 = 0$ A.0B.1C.2D.3

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Hint: We will simplify the terms of the given equation and then we will use the properties of logarithmic functions for further simplification and then we will substitute ln x=t for simplicity of the cubic equation so formed after simplification because if we carry forward the calculations using the cubic equation in terms of lnx, it will become more complex for further solution.

We are given the equation ${4^{{{\log }_2}\log x}} - 1 + {\ln ^3}x - 3{\ln ^2}x - 5\ln x + 7 = 0$
We will simplify the first term of this equation using the property of logarithmic functions: $a\log x = {\left( {\log x} \right)^a}$
$\Rightarrow {4^{{{\log }_2}\log x}} = {2^{2\left( {{{\log }_2}\log x} \right)}} \\ \Rightarrow {2^{{{\log }_2}{{\left( {\log x} \right)}^2}}} \\$
Now, we know another property of logarithmic functions which is ${a^{{{\log }_a}x}} = x$.
Using above mentioned property in the first term(simplified), we get
$\Rightarrow {4^{{{\log }_2}\log x}} = \log {x^2}$
Substituting the value of the first term in the given equation, we get:
${\ln ^2}x - 1 + {\ln ^3}x - 3{\ln ^2}x - 5\ln x + 7 = 0 \\ \Rightarrow {\ln ^3}x - 2{\ln ^2}x - 5\ln x + 6 = 0 \\$
We get a cubic equation in terms of lnx. For simplification, let us suppose lnx= m.
Substituting the value of ln x = m in the cubic equation, we get
${m^3} - 2{m^2} - 5m + 6 = 0$
Now for the values which will satisfy the equation, we can tell by observation that 1 is a factor of the above mentioned cubic equation.
$\therefore {1^3} - 2 \times {1^2} - 5 \times 1 + 6 = 0 \\ \Rightarrow 1 - 2 - 5 + 6 = 0 \\$
1 satisfies the cubic equation. Therefore, m-1 will be a factor of this cubic equation.
Now, dividing the cubic equation by the factor m – 1, we get
$\dfrac{{{m^3} - 2{m^2} - 5m + 6}}{{m - 1}} = {m^2} - m - 6$
Therefore, ${m^2} - m - 6 = 0$ will also be a factor of the cubic equation.
We will factorise the factor ${m^2} - m - 6 = 0$
$\Rightarrow {m^2} - 3m + 2m - 6 = 0 \\ \Rightarrow m\left( {m - 3} \right) + 2\left( {m - 3} \right) = 0 \\ \Rightarrow m = 3 \, and \, m = - 2 \\$
We have supposed m= lnx, therefore substituting values back of m= 1, m= -2 and m= 3, we get
$\ln x = 1 \Rightarrow x = {e^1} \\ \ln x = - 2 \Rightarrow x = {e^{ - 2}} \\ \ln x = 3 \Rightarrow x = {e^3} \\$
Therefore, we get in total three values which satisfy the given equation.

Hence, option(D) is correct.

Note: In such questions, one may get confused with the use of properties of logarithmic functions and also with simplification of the factors. Logarithm properties and solutions of cubic should be known to students to solve this type of problem.