Question

Sketch the graph for $y={\log }_{\frac{1}{4}}(x-\frac{1}{4})+\frac{1}{2}{\log }_4(16x^2-8x+1)$

Answer 1

Here, $y={\log }_{\frac{1}{4}}(x-\frac{1}{4})+\frac{1}{2}{\log }_4(16x^2-8x+1)$
$\Rightarrow y=-{\log }_4(x-\frac{1}{4})+\frac{1}{2}{\log }_4{(4x-1)}^2$
$\Rightarrow y=-{\log }_4(x-\frac{1}{4})+\frac{1}{2}{\log }_4{4}^2+\frac{2}{2}{\log }_4(x-\frac{1}{4})$
$\Rightarrow y=-{\log }_4(x-\frac{1}{4})+{\log }_4(x-\frac{1}{4})+\frac{2}{2}{\log }_{4}4$
As ${\log }_{a}x$ exists only when $a,x>0$ and $a\ne 1$
$\Rightarrow y=1,$ whenever $(x-\frac{1}{4})>0$
Thus,$y={\log }_{\frac{1}{4}}(x-\frac{1}{4})+\frac{1}{2}{\log }_4{(4x-1)}^2$
$\Rightarrow$ Domain$\in (\frac{1}{4},\infty )$
Range$\in \left\{1\right\}$
Thus, the graph is shown below
Thus, the graph for
$y={\log }_{\frac{1}{4}}(x-\frac{1}{4})+\frac{1}{2}{\log }_4{(4x-1)}^2$