solve the equation of $\log(\frac{x}{2})=\frac{2}{\log x}$

Question

Solve the equation of $\log(\frac{x}{2})=\frac{2}{\log x}.$

I am not sure how can I solve it! I have tried to use other way which there are $2$ answers and I think it should not be .please help me !!!

Answer 1

Notice: $log(x/2) = log(x)-log(2)$

$(log(x)-log(2))/1 = 2/log(x)$

$(log(x)-log(2))*log(x) = 2$

Let $y = log(x)$

$(y-log(2))*y = 2 $

$y^2-log(2)*y -2 =0 $

$y= \frac{-(-log(2))\pm sqrt{(-log(2))^2 - 4*1*(-2)}}{2*1}$

$log(x) = \frac{-(-log(2))\pm \sqrt{(-log(2))^2 - 4*1*(-2)}}{2*1}$

$log(x) = \frac{log(2)\pm \sqrt{8+\log^2(2)}}{2}$

$log(x) = log(\sqrt{2}) \pm \frac{\sqrt{8+\log^2(2)}}{2}$

$x = e^{log(\sqrt{2}) \pm \frac{\sqrt{8+\log^2(2)}}{2}}$

$x = \sqrt{2}*e^{\pm\frac{\sqrt{(8+\log^2(2))}}{2}}$

Answer 2

We can start by using the identity $\log(a/b)=\log(a)-\log(b)$.
We get $\log(x)-\log(2)=\frac{2}{\log(x)}$
After this we say that log(x)=y,
$y-\log 2 = \frac2y$, and multiply by y on both sides, and we get

$y^2-y\log(2)=2$
and we subtract 2 from both side and get a quadratic equation and solve it,and we get that the 2 roots are $y_{1,2}=\frac{log(2)\pm\sqrt{(log2)^2+8}}{2}$

and finally going back to $x$ (by saying $x=e^y$) we get that the 2 solutions are

$x_{1,2}=e^{\left(\frac{log(2)\pm\sqrt{(log2)^2+8}}{2}\right)}$

thanks for reading and goodbye.