Question

If ‘a’ and ‘b’ be the rational and irrational solution respectively of equation
${{\log }^{2}}(100x)+{{\log }^{2}}\left( 10x \right)=14+\log \left( \dfrac{1}{x} \right)\text{ then value of }a+{{b}^{{}^{-2}/{}_{9}}}\text{ is}$

1. 10
2. 20
3. 30
4. 40

Answer 1

use the sum and divisor property of log functions to create the whole equation in single variable to find the value of x. the logarithmic function of sum and subtraction respectively is
$\Rightarrow \text{ }\log \text{a}+\log \text{b}=\log \left( ab \right)$
$\Rightarrow \text{ }\log a-\text{log }b=\log \left( {}^{a}/{}_{b} \right)$

Complete step by step solution: Let’s begin with, what the expression is given it’s
${{\log }^{2}}(100x)+{{\log }^{2}}(10x)=14+\log \left( \dfrac{1}{2} \right)$
If we check the expression, we are unable to get a whole 10g as a single variable, means 10g (100x) as one variable directly. we need to convert this into single variable.
As we know the sum property of logarithmic function work as
$\log a+\log b=\log ab$
We can write log $\left( 100x \right)=\log 100+\log \text{ }x$
Similarly, $\log \left( 10\text{ }x \right)=\log 10+\log x$
and we know that subtraction property works as
$\Rightarrow \log a\log b=\log \left( \dfrac{a}{b} \right)$
If we try to use this all property in the equation, we get

& \Rightarrow \text{ }{{\left( \log 100x \right)}^{2}}+{{\left( \log 10x \right)}^{2}}=14+\log \left( {}^{1}/{}_{x} \right) \\\ & \Rightarrow \text{ }{{\left( \log 100+\log x \right)}^{2}}+{{\left( \log 10+\log x \right)}^{2}}=14+\left( 10g\text{ }1-10g\text{ }x \right) \\\ \end{aligned}$$ Here let’s assume that, the base of 10g is 10, then the change in expression will be $$\Rightarrow {{\left( 2+\log x \right)}^{2}}+{{\left( 1+\log x \right)}^{2}}=14+0-\log x$$ As we can see, $$\log x$$ is the unknown variable only. we can assume $$\log x$$ as a variable d. so that expression changes to $$\begin{aligned} & {{\left( 2+d \right)}^{2}}+{{\left( 1+d \right)}^{2}}=14-d \\\ & \Rightarrow \text{ }4+4d+{{d}^{2}}+1+2d+{{a}^{2}}=14-d \\\ & \Rightarrow \text{ }2{{d}^{2}}+7d-9=0 \\\ \end{aligned}$$ Using quadratic formula, we get $$\begin{aligned} & \dfrac{d=-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{=}\dfrac{-7\pm \sqrt{49-4\times 2\times \left( -9 \right)}}{2\times 2} \\\ & \Rightarrow \text{ }d=\dfrac{-7\pm \sqrt{121}}{4}=\dfrac{-7\pm 11}{4} \\\ \end{aligned}$$ So, d can be $$\dfrac{-7+11}{4}v=1\text{ and }\dfrac{-7-11}{4}=\dfrac{-18}{4}=\dfrac{-9}{2}$$ Hence, we have $$d=\log x$$ we need to find the value pf $$x$$ So, for $$d\text{ }=\text{ }1$$ we get $$\begin{aligned} & \Rightarrow \text{ }\log 10x\text{ }=1 \\\ & \Rightarrow \text{ }x=10 \\\ \end{aligned}$$ and for $$d={}^{-9}/{}_{2}$$ we get $$\begin{aligned} & \Rightarrow \log 10x={}^{-9}/{}_{2} \\\ & \Rightarrow x={{10}^{{}^{-9}/{}_{2}}} \\\ \end{aligned}$$ Hence, we get two values of x and we need the sum in the form of $$a+{{b}^{{}^{-2}/{}_{9}}}$$. As we can see the irrational form should assign to b for getting a whole no. as an output. So, we assign $$a=10\text{ and }b={{10}^{{}^{-9}/{}_{2}}}$$. So, we get $$a+{{b}^{{}^{-2}/{}_{9}}}=10+{{\left( {{10}^{{}^{-9}/{}_{2}}} \right)}^{{}^{\times -2}/{}_{9}}}=10+10=20$$ **Hence, option b is the correct answer.** **Note:** The logarithmic properties are used to generate a common variable. like $${{\log }_{b}}\text{ }a\text{ }=\text{ }\dfrac{{{\log }_{e}}a}{{{\log }_{e}}a}\text{ }=\text{ }\dfrac{{{\log }_{10}}a}{{{\log }_{10}}a}$$. In logarithm function as a variable, always try to get the similar value in logarithmic function, to treat it as a variable and form equation to obtain its roots.