Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.
My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\sqrt{2}}(x)=\frac{\ln(x)}{\ln(\sqrt{2})}$$
then, we can re-write the equation as $$\left(\frac{\ln(x)}{\ln(2)}\right)^{2}-\left( \frac{\ln(x)}{\ln(\sqrt{2})}\right)^{2}=a-\sqrt{a+\frac{\ln(x)}{\ln(2)}}$$
Let $t:=\ln(x)$ so the equation above we can re-write in terms of $t$ as $$\left(\frac{1}{\ln(2)}t\right)^{2}-\left( \frac{1}{\ln(\sqrt{2})}t\right)^{2}=a-\sqrt{a+\frac{1}{\ln(2)}t}$$
Then, the idea that I was thinking is squaring both sides of the equation, but there appears an equation of degree 4 that is difficult to solve. How could I continue from there? or maybe a simpler approach?
New approach: Using the point out of @Ross Millikan and since that $$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt{2})}=2\log_{2}(x)$$
and let $t:=\log_{2}(x)$ so the equation can be we-write in terms of $t$ as $$t^{2}-2t=a-\sqrt{a+t} \implies t^{2}-2t\color{blue}{+1}=a-\sqrt{a+t}\color{blue}{+1} \implies (t-1)^{2}=a+1-\sqrt{a+t}$$
or maybe $$ t^{2}-2t=a-\sqrt{a+t} \overset{y^{2}=a+t}{\implies} (y^{2}-a)^{2}-2(y^{2}-a)=a-y^{2}$$
solving a little the expression above we can see that $$y^{4}-2y^{2}a+a^{2}-2y^{2}+2a-a+y^{2}=0$$
how can I solve this?
Best Answer
Edit: @mathlove noticed problems with my initial solution. My effort to resolve those problems ended in a major revision of the solution. Please comment if you find mistakes in this new version.
Let's take off from where you calculated that $t^2-2t=a-\sqrt{a+t}$ . Our aim is to find a value (or values) for $a$ such that the above equation has exactly 2 real answers for $t$. Let's also keep in mind the implicit constraint that $$a+t \ge 0 \Rightarrow t \ge -a \qquad(1)$$ This is because the given problem includes $\sqrt{a+t}$ . An important implication of (1) is that if $a$ is negative then $t$ should be positive, and if $t$ is negative then $a$ should be positive.
Another constraint that is derived from our starting equation is that $$t^2-2t \le a \qquad (2)$$ where the equality holds when $a+t=0$. Besides, (2) means that if $a < 0$ then $t^2-2t=t(t-2) < 0$ , which implies that $0 < t < 2$.
And two other constraints, which we can derive from (2) are $$t^2-2t+1 = (t-1)^2 \le a+1 \qquad (3)$$ $$0 \le (t-1)^2 \le a+1 \Longrightarrow a \ge -1 \qquad (4)$$
Alright, here we go: $$t^2-2t=a-\sqrt{a+t}$$ $$t^2-t=a+t-\sqrt{a+t}$$ $$t^2-t+\frac{1}{4}=a+t-\sqrt{a+t}+\frac{1}{4}$$ $$(t-\frac{1}{2})^2 = (\sqrt{a+t}-\frac{1}{2})^2$$ $$t-\frac{1}{2} = \pm(\sqrt{a+t}-\frac{1}{2})$$ So we have two (quadratic) equations $$t-\frac{1}{2} = +(\sqrt{a+t}-\frac{1}{2}) \Rightarrow t^2 = a+t \qquad (5)$$ and $$t-\frac{1}{2} = -(\sqrt{a+t}-\frac{1}{2}) \Rightarrow (t-1)^2 = a+t \qquad (6)$$ Let's look more carefully at (5). The solutions to this quadratic equation are $$t_1 = \frac{1 + \sqrt{1+4a}}{2} , t_2 = \frac{1 - \sqrt{1+4a}}{2}$$ For these solutions to exist, we should have $a \ge -\frac{1}{4}$.
Consider $t_1$. There seems to be no problem with $t_1$ for $-\frac{1}{4} \le a$ . In this range of $a$ , $t_1$ seems to satisfy constraints (1),(2),(3) and (4).
Now consider $t_2$. If $a > 0$ then by constraint (2) we should have: $$t_2^2 - 2t_2 \le a$$ $$(\frac{1 - \sqrt{1+4a}}{2})^2 - 2(\frac{1 - \sqrt{1+4a}}{2})= \frac{-1+2a+\sqrt{1+4a}}{2} \le a$$ $$-1+2a+\sqrt{1+4a} \le 2a$$ $$\sqrt{1+4a} \le 1$$ $$1+4a \le 1$$ $$a \le 0$$ which is a contradiction with our assumption of $a > 0$. So, $t_2$ does not exist for $a > 0$. There seems to be no problem with $t_2$ for $-\frac{1}{4} \le a \le 0$ . In this range of $a$ , $t_2$ seems to satisfy constraints (1),(2),(3) and (4).
Note that if $a = -\frac{1}{4}$ then $t_1=t_2=\frac{1}{2}$.
Let's now look at (6). The solutions to this quadratic equation are $$t_3 = \frac{3 + \sqrt{5+4a}}{2} , t_4 = \frac{3 - \sqrt{5+4a}}{2}$$ For these solutions to exist, we should have $a \ge -\frac{5}{4}$ , but we also have the stricter constraint (4), which requires that $a \ge -1$.
Consider $t_3$. If $a > 0$ then $t_3$ violates constraint (3). So, $t_3$ does not exist for $a > 0$. On the other hand, if $-1 < a \le 0$ then by constraint (1) t should be non-negative and by the implication of constraint (2) we should have $0 \le t \le 2$. But applying this range of $a$ in the calculation of $t_3$ results in $2 < t_3$, which is in contradiction with constraint (2). Finally, note that if $a=-1$ then $t_3=2$ , but this result violates constraint (3). Therefore, it seems that $t_3$ is not a solution to the equation for any value of $a$.
Now consider $t_4$. It seems that for $a \ge -1$ , $t_4$ satisfies constraints (1),(2),(3) and (4).
Note that although in the formula we get $t_3=t_4=\frac{3}{2}$ for $a=-\frac{5}{4}$ , these are not solutions of the main equation, because the value of $a=-\frac{5}{4}$ violates constraint (4).
Now let us see in what conditions $t_1$ or $t_2$ might be equal to $t_4$. From $t_1 = t_4$ it follows that $a=-\frac{1}{4}$. Also, from $t_2 = t_4$ it follows that $a=-\frac{1}{4}$. So at $a=-\frac{1}{4}$ , $t_1=t_2=t_4=\frac{1}{2}$.
Summing up our above analyses, we can see that
$t_1$ exists for $-\frac{1}{4} \le a$
$t_2$ exists for $-\frac{1}{4} \le a \le 0$
$t_3$ does not exist
$t_4$ exists for $-1 \le a$
Therefore, for $a > 0$ , the equation has exactly two distinct solutions ($t_1 , t_4$).