algebraic equation with powers

Question

The original statement of the problem : Solve in $\mathbb R$ the following equation : $$ a^{log_b x^2 } + a^{log_x b^2 } = a^{1+log_b x } + b^{1+log_x b } $$ where $a,b>0$ and $b \neq 1$

A more general statement of the problem : Find all real solutions t , to the equation $ a^{2t} + a^{\frac{2}{t}} = a^{1+t} + a^{1+ \frac{1}{t}} $ where a is a strictly positive real constant.

EDIT I hope I did not complicate the problem by making these transformations to the second equation, I am especially interested in solving the original one, so that's more important.

My approach: First of all, if a = 1, it is clear that any real t verifies the given equation. In the following, I assumed that a > 1 to simplify, and then take the case when 0<a<1, analogously. I tried to prove that t = 1 is the only solution of the equation for any a different from 1. I transformed the equation into this

$a^{t}(a^{t}-a) + a^{\frac{1}{t}}(a^{\frac{1}{t}}-a)=0$

where I was going to use some inequalities and prove that the left side is strictly greater for t different from 1 but I got stuck.

Any and all proofs will be helpful. Thanks a lot!

Answer 1

Solution for $a > 1$:

As you mentioned, we begin with $a > 1$. Let us consider $t \leq 0$. Clearly $a^t - a < 0$ and $a^{1/t} - a < 0$ so the equation clearly does not have a solution for $t \leq 0$. So let us focus on $t > 0$. Define $f(t) = a^t(a^t - a)$. It is straightforward to note that $f$ is a strictly increasing, convex function. We are interested in the solution of $$ g(t) = f(t) + f\left(\frac{1}{t} \right).$$ For any $p > 0$, if $t' > 0$ is a solution to $g(t) = p$, then so is $1/t'$. Since $1/x$ is a convex function and $f$ is increasing, $g$ is a strictly convex function. To show that let $h(t) = 1/t$ for simplicity. For any $\lambda \in [0,1]$ and $x,y > 0$, we have, \begin{align*} g(\lambda x + (1-\lambda)y) & = f(h(\lambda x + (1-\lambda)y)) + f(\lambda x + (1-\lambda)y) \\ & \leq f(h(\lambda x + (1-\lambda)y)) + \lambda f(x) + (1-\lambda)f(y) \\ & < \lambda f(h(x)) + (1-\lambda)f(h(y)) + \lambda f(x) + (1-\lambda)f(y) \\ & < \lambda g(x) + (1-\lambda)g(y). \end{align*} In second and fourth line, we used convexity of $f$ and in third line we used convexity of $h$ and that $f$ is strictly increasing. Thus, $g$ is strictly convex. This implies that there can be at most $2$ solutions to $g(t) = p$ for any $p \in \mathbb{R}$. These, in fact, are $t'$ and $1/t'$ as mentioned above.

We know that $t^{\star} = 1$ is a solution to $g(t) = 0$. The only other solution has to be $1/t^{\star} = 1$. Thus, $t = 1$ is the only solution.

For $a < 1$, once again we can show that there no solutions for $t \leq 0$ as $a^{t} - a > 0$ and $a^{1/t} - a > 0$. Thus, we can again focus on $t \geq 0$. Now, some initial results suggest that there are $3$ solutions for certain values of $a \in (0,1)$. If you want I can add some additional discussion as to why there will be exactly $3$ solution, but that definitely requires me to use derivatives.

Answer 2

Unfortunately this answer is not quite finished, but I have to run, so posting it here to save it and/or have others finish it.

Let $a\in\Bbb{R}_{>0}$ be a positive real number, and let $t\in\Bbb{R}$ be a real number such that $$a^{2t} + a^{\frac2t} = a^{1+t} + a^{1+ \frac1t}.\tag{0}$$ First note that $t=1$ is a solution for any $a\in\Bbb{R}_{>0}$, so we will only consider $t\neq1$ from here on. Similarly, for $a=1$ any $t\in\Bbb{R}$ is a solution, so we will only consider $a\neq1$ from here on. We will also not consider $t=0$, as the original equation is not defined at this point.

Let $b:=a^{\tfrac1t}$ so that $a=b^t$ and hence $$b^{2t^2}+b^2=b^{t^2+t}+b^{t+1}.$$ Rearranging, dividing by $b^2$ and factoring the exponents, we find that $$b^{2(t+1)(t-1)}-b^{(t+2)(t-1)}-b^{t-1}+1=0,$$ and hence for $c:=b^{t-1}$ that $$(c^t)^2-c^t=\frac{c-1}{c^2}.\tag{1}$$ Note that $c=a^{\frac{t-1}{t}}$ is again a positive real number, and that conversely, if $c\in\Bbb{R}_{>0}$ is a positive real number and $t\in\Bbb{R}$ a real number satisfying $(1)$, then $a:=c^{\frac{t}{t-1}}$ and $t$ satisfy the original equation $(0)$. So it suffices to find all solutions to $(1)$, where we only consider solutions with $t\neq1$ and $a\neq1$, and hence $c\neq1$.

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Let $C\in\Bbb{R}_{>0}$ be a positive real constant different from $1$, and consider the real-valued function $$f:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}:\ x\ \longmapsto\ (C^x)^2-C^x.$$ It is easy to see that $\lim_{x\to\infty}f(x)=0$ if $C<1$ With some basic calculus we find that $$f'(x)=C^x(2C^x-1)\ln C,$$ from which it follows that $f'(x)=0$ if and only if $x=-\frac{\ln 2}{\ln C}$, and we readily see that $-\tfrac14$ is the unique minimum of $f$ at $x_0:=-\frac{\ln 2}{\ln C}$. Clearly $f(0)=0$ regardless of the value of $C$, and so we see that:

• If $C>1$ then $\lim_{x\to-\infty}f(x)=0$. It follows that $f(x)<0$ for all $x<0$. Because $\frac{C-1}{C^2}>0$ we see that there is no solution to $(1)$ with $c>1$ and $t\leq0$.
• If $C<1$ then $\lim_{x\to-\infty}=\infty$. It follows that $f(x)>0$ for all $x<0$. Because $\frac{C-1}{C^2}<0$ we see that there is no solution to $(1)$ where $c<1$ and $t\leq0$.

In conclusion, if $(c,t)$ is a solution to $(1)$ with $c\neq1$ then $t>0$. Correspondingly, if $(a,t)$ is a solution to $(0)$ with $a\neq1$ then $t>0$.

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Now viewing $(1)$ as a quadratic in $c^t$ we find that $$c^t=\frac{1\pm\sqrt{1+4\frac{c-1}{c^2}}}{2}.$$ In particular we see that there is no solution if $1+4\frac{c-1}{c^2}<0$, or equivalently $c<2\sqrt{2}-2$.

If $c>1$ then $\sqrt{1+4\frac{c-1}{c^2}}>1$ and so we must have the $+$-sign, because of course $c^t>0$. Similarly, if $c<1$ then $c^t<1$ because we saw that $t>0$, and so we must have the $-$-sign. With this in mind we find that $$t(c)=\frac{\ln\tfrac12\left(1\pm\sqrt{1+4\frac{c-1}{c^2}}\right)}{\ln c},$$ and correspondingly $a=c^{\frac{t(c)}{t(c)-1}}$.

For $c>1$ it is not difficult to check that $t(c)<1$ and hence that $a=c^{\frac{t}{t-1}}<1$, and that $t(c)$ is strictly decreasing. For $c<1$ we already noted that $t(c)$ is not defined whenever $c<2\sqrt{2}-2$. For $c\geq2\sqrt{2}-2$ we see that $t(c)$ is strictly increasing with $t(2\sqrt{2}-2)\approx3.6825\ldots$. Again we see that $a=c^{\frac{t(c)}{t(c)-1}}<1$.

In particular, we see that there are no solutions with $a>1$.

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Some unsorted ideas.

Now suppose $C<1$. Then $f(0)=0$ and $\lim_{x\to\infty}f(x)=0$, and the unique extremum of $f$ is $f(x_0)=-\tfrac14$ at $x_0:=-\frac{\ln 2}{\ln C}$. This already shows that we cannot have $$(C^x)^2-C^x=\frac{C-1}{C^2},$$ if $\frac{C-1}{C^2}<-\frac14$, or equivalently, if $C<2\sqrt{2}-2\approx0.8284\ldots$. But if $C\geq2\sqrt{2}-2$ then

Now suppose $C>1$. Then $f(0)=0$ and $\lim_{x\to\infty}f(x)=\infty$, and $f'(x)>0$ for all $x>0$, and so there is precisely one real number $t\in\Bbb{R}$ such that $$(C^t)^2-C^t=\frac{C-1}{C^2}.\tag{2}$$ Denote this value by $t(C)$. Then we have seen before that $t(C)>0$. Viewing the right hand side of $(2)$ as a function of $C$, it is easily verified to have a maximum of $\tfrac14$ at $C=2$. This means $$(C^{t(C)})^2-C^{t(C)}=\frac{C-1}{C^2}\leq\frac14,$$ which already tells us that $$t(C)\leq\frac{\ln\left(\tfrac{1+\sqrt{2}}{2}\right)}{\ln C}.$$ In particular, for $C>\tfrac{1+\sqrt{2}}{2}$ we see that $t(C)<1$ and hence that $a:=C^{\frac{t(C)}{t(C)-1}}<1$. For $C\leq\tfrac{1+\sqrt{2}}{2}$ we find that $$\frac{C-1}{C^2}\leq-14+10\sqrt{2}\approx0.14214\ldots,$$ which as before tells us that $$t(C)\leq\frac{\ln\frac12\sqrt{40\sqrt{2}-55}}{\ln C},$$ and as before, this in turn for $C>\frac12\sqrt{40\sqrt{2}-55}$ implies that $a:=C^{\frac{t(C)}{t(C)-1}}<1$.

[Iterating this process will show that for every $C>1$ we get $t(C)<1$ and hence $a:=C^{\tfrac{t(C)}{t(C)-1}}<1$. I have not made this rigorous yet, and there must be a nicer way to go about showing this.]