Find the values of \(y\) such that \(\log _{,} 4+\log _{4} \frac{1}{y}=\left(\frac{2}{3}\right)^{-1}\)
A. \(1.2\) B. \(\frac{1}{2} , 3 \) C. \(\frac{1}{3} ,16 \) D. \(2 , \frac{1}{16}\) Correct Answer: DExplanation2 and \(1 / 16\) To solve the equation; \begin{array}{l} \log _{1} 4+\log _{4} \frac{1}{y}=\left(\frac{2}{3}\right)^{-1} \\ \text { since }\left(\frac{2}{3}\right)^{-1}=\frac{1}{2 / 3}=\frac{3}{2} \\ \Rightarrow \log _{1} 4+\log _{4} \frac{1}{y}=\frac{3}{2} \\ \Rightarrow \log _{1} 4+\log _{4} y^{-1}=\frac{3}{2} \\ \log _{1} 4-\log _{4} y=\frac{3}{2} \\ \text { Note that } \log _{4} y=\frac{1}{\log _{1} 4} \\ \Rightarrow \log _{1} 4-\frac{1}{\log _{,} 4}=\frac{3}{2} \end{array} let \(\log _{y} 4=x\) \(\frac{x}{1}-\frac{1}{x}=\frac{3}{2}\) we take the LC.M. \(\frac{x^{2}-1}{x}=\frac{3}{2}: x^{2}-1=\frac{3 x}{2}\) muliiply through by 2 To solve the equation or \(2 x^{2}-3 x-2=0\) \(2 x^{2}-4 x+x-2=0\) \(2 x(x-2)+1(x-2)=0\) \((2 x+1)(x-2)=0\) \(2 x+1=0, x-2=0\) \(x=-1 / 2\) or \(x=2\) but \(x=\log _{2} 4\) \(\Rightarrow-\frac{1}{2}=\log _{x} 4\) or \(\log _{x} 4=-\frac{1}{2}\) changing to index form \(4=y \div\) \(\Rightarrow 4=\frac{1}{y^{12}}\) : \(4=\frac{1}{\sqrt{y}}\) or \(4 \sqrt{y}=1\) \(\sqrt{y}=\frac{1}{4}\) square both sides \(y=\frac{1}{16}\) afso, when \(x=2, \log , 4=2\) changing to index form. we have \(4=y^{2} \text { or } y^{2}=4, y=\pm \sqrt{4}=\pm 2\)
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