Question

If $4^{{\log }_{10}[x+1]}-6^{{\log }_{10}x}-2\cdot 3^{{\log }_{10}[x^2+2]}=0$ then the value of $\frac{1}{20x}$ is

Answer 1

Correct Answer: 5

Explanation:
Consider,$4^{{\log }_{10}(x)+1}-6^{{\log }_{10}x}-2\cdot 3^{{\log }_{10}\left(x^2\right)+2}=0$$\Rightarrow {\left[2^2\right]}^{{\log }_{10}(x)+1}-[2\cdot 3{]}^{{\log }_{10}x}-2\cdot 3^{{\log }_{10}{x}^2}\cdot 3^2=0$$\Rightarrow 2^{2{\log }_{10}x+2}-2^{{\log }_{10}x}\cdot 3^{{\log }_{10}x}-2\cdot 3^{{\log }_{10}{x}^2}\cdot 3^2=0$[Using properties of logarithmic function-3 ]$\Rightarrow 4\cdot 2^{{\log }_{10}{x}^2}-2^{{\log }_{10}x}\cdot 3^{{\log }_{10}x}-18\cdot 3^{2{\log }_{10}x}=0$Now, substitute $2^{{\log }_{10}x}=A$ and $3^{{\log }_{10}x}=B,$ we get$4A^2+AB-18B^2=0\text{ [Using middle term split }]$$\Rightarrow (4A-9B)(A+2B)=0$But, $A+2B\ne 0$ as $A>0$ AND $B>0$$\Rightarrow 4A=9B$$\Rightarrow 4\cdot 2^{{\log }_{10}x}=9\cdot 3^{{\log }_{10}x}$$\Rightarrow \frac{4}{9}={\left(\frac{3}{2}\right)}^{{\log }_{10}x}$$\Rightarrow {\left(\frac{3}{2}\right)}^{-2}={\left(\frac{3}{2}\right)}^{{\log }_{10}x}$$\Rightarrow {\log }_{10}x=-2$[Using properties of logarithmic function-10]$\Rightarrow x=\frac{1}{100}$Thus, $\frac{1}{20x}=5$Hence, the answer is 5.00.