Question

If $x=\alpha$ is the solution of the equation $\left|2+{\log }_{2}7x\right|-{\log }_2(x-1)=5$, then find the value of $(65{)}^{\frac{1}{3}{\log }_{{\alpha }^2+1^{\alpha }}}$.

Answer 1

Answer: 2

For ${\log }_2(x-1)$ to be defined $x$ should be greater than 1 and $2+{\log }_{2}7x>0$
$\therefore 2+{\log }_{2}7x-{\log }_2(x-1)=5$
$\Rightarrow {\log }_2\left(\frac{7x}{x-1}\right)=3\Rightarrow \frac{7x}{x-1}=8\Rightarrow 7x=8x-8$
$\Rightarrow x=8=\alpha$
Now,
$(65{)}^{\frac{1}{3}}$
)${}^{\frac{1}{3}{\log }_{{\alpha }^2+1^a}}=(65{)}^{\frac{1}{3}{\log }_{65}8}=8^{\frac{1}{3}}=2$