Question

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

Answer 1

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