Prove (or disprove): If $a,b,c,d$ are positive real numbers with $a\times b=c\times d$, then the only solutions for $x$ in the equation $$\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{c+d}$$ are $x = \pm 1$.

Other than the obvious $a=b=c=d$ solution.

  • $\begingroup$ Consider taking the limit as $x \rightarrow \pm \infty$. This would prove the only solution is $\pm1$ $\endgroup$
    – MathsPro
    Apr 19 '15 at 14:22
  • $\begingroup$ Can you elaborate on it MathsPro? $\endgroup$
    – GohP.iHan
    Apr 19 '15 at 16:58

Clearly, if $a = c, b = d$ or $a = d, b = c$, then all $x$ satisfy the equation. Hence the statement is false. However, one can show that this is the only exception, namely, if $a,b,c,d > 0$ such that $ab = cd$ and $\{a,b\}\neq\{c,d\}$, then $x = \pm 1$ is the only solution to $$\frac{a^x+b^x}{a+b} = \frac{c^x+d^x}{c+d}.$$

Suppose $ab = cd = s^2$. Without loss of generality, we may assume $s = 1$ since otherwise we can divide $a,b,c,d$ by $s$ and the solution set will not change. We may also assume $a \ge b$ and $c \ge d$. Now $b = 1/a, d = 1/c$ and $a,c \ge 1$.

Consider $f\colon [0,\infty)\times \mathbb{R}\to\mathbb{R}$ defined by $$f(z, x) = \frac{e^{zx} + e^{-zx}}{e^{z} + e^{-z}}.$$

Claim $f(z,x)$ is decreasing in $z$ if $|x| < 1$ and is increasing in $z$ if $|x| > 1$. Therefore $f$ is injective in $z$ if $x\neq \pm 1$.

Suppose the claim holds. We know that $f(\log a,x)=f(\log c,x)$ implies either $x = \pm 1$ or $a = c$.

Proof of Claim Observe that $$\partial_z f(z,x) = (-1 + x)\left(e^{(1+x)z}-e^{-(1+x)z}\right) + (1+x)\left(e^{(x-1)z}-e^{(1-x)z}\right).$$ If $|x| < 1$, then all terms above are negative and so $f$ is decreasing in $z$. If $|x| > 1$, then all terms above are positive, and so $f$ is increasing in $z$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.