The Special Mean

First, solve the equation: $\log_{a}{\lambda}=\log_{\lambda}{b}$ $\frac{\ln{\lambda}}{\ln{a}}=\frac{\ln{b}}{\ln{\lambda}}$ ${(\ln{\lambda})}^2=\ln{a}\ln{b}$ $\ln{\lambda}=±\sqrt{\ln{a}\ln{b}}$ $\lambda=e^{±\sqrt{\ln{a}\ln{b}}}$ $\lambda=a^{±\sqrt{\frac{\ln{b}}{\ln{a}}}}=b^{±\sqrt{\frac{\ln{a}}{\ln{b}}}}$

There are two possible values for the special mean.

The second equation $\log_{a}{\lambda}=\log_{\lambda}{b}$ will return the same results for $\lambda$ . This means that $a$ and $b$ are interchangeable. Assume then that $a≤b$ .

It can be easily shown that $a≤a^{\sqrt{\frac{\ln{b}}{\ln{a}}}}=\lambda$ and $b≥b^{\sqrt{\frac{\ln{a}}{\ln{b}}}}=\lambda$ . Thus, one of the special mean is always between (inclusive) $a$ and $b$ .

However, this is not true for the second possible value for the special mean. In fact, this value is only in between $a$ and $b$ when $a=b=1$ . Unfortunately, in this case, the original definition does not hold, as one cannot take the logarithm at base $1$ .

Thus, the answer is $\text{No}$ .

If $\log_a \lambda = \log_{\lambda} b$ then $\frac{ \ln \lambda}{ \ln a} = \frac{ \ln b}{ \ln \lambda}$ and solving for $\lambda$ , $\lambda = e^{\sqrt{\ln b \ln a}}$ . After having this expression and playing around with it an easy way to construct a counter example arises by asking a simple question: Can really small $a,b$ yield really big $\lambda$ ? The answer is yes, and I will just share the first counter example I constructed even though it is easy to produce other ones. Let $a= e^{-e^{\frac{2}{4}}}$ and $b = e^{-e^{\frac{6}{4}}}$ . It is clear that these numbers are really small, both obviously smaller than 1. Then we can compute $\lambda = e^{\sqrt{-e^{\frac{2}{4}} * -e^{\frac{6}{4}}}} = e^{\sqrt{e^2}} = e^e$ and this number is obviously bigger than 1 and therefore it cannot be between $a$ and $b$ .

To show the negation of the question it just suffices a counterexample. Set a=2 and b=16 ad hoc and you get lambda=1/4 that is not between a and b (included).

The situation is rather similar to the definition of the geometric mean in the way proposed. If $a$ and $b$ then $\frac b{-\sqrt{ab}} = -\sqrt{\frac{b}{a}} = \frac{-\sqrt{ab}}a,$ which proves that $-\sqrt{ab}$ is "a geometric mean" of $a$ and $b$ , even though it is less than both of them.

Now replace $a$ by $\log a$ and $b$ by $\log b$ .

That is, for $a,b > 1$ , define $\log \lambda = -\sqrt{\log a\cdot \log b}$ ; then $\log_a \lambda = \frac{\log \lambda}{\log a} = \frac{-\sqrt{\log a\cdot \log b}}{\log a} = -\sqrt{\frac{\log b}{\log a}} = \frac{\log b}{-\sqrt{\log a\cdot \log b}} = \frac{\log b}{\log \lambda} = \log_\lambda b.$ Thus $\lambda$ is "a special mean" of $a$ and $b$ , yet $\lambda < 1 < a,b$ .

Let us suppose that $0 < a \le b$ . The condition to be a special mean tells us that $(\ln \lambda)^2 = \ln a\, \ln b$ . Thus we must have $\ln a \, \ln b \ge 0$ and then $\ln \lambda \; = \; \pm \sqrt{\ln a\, \ln b}$ Thus there are two possible special means, and one is the inverse of the other. If both of these special means lie between $a$ and $b$ , then (multiplying them together) $1$ must lie between $a^2$ and $b^2$ , so that $a \le 1 \le b$ . If $a < 1 < b$ then $\ln a\, \ln b <0$ , and so no special mean exists. Thus we deduce that at least one of $a, \,b$ must be equal to $1$ . But this is not possible, since the definition of the special mean requires us to be able to define logarithms to bases $a$ and $b$ , and there is no logarithm to base $1$ .

Thus it is never the case that all special means (that exist) lie between $a$ and $b$ .

The way I solved it is by assuming $a=b\neq (0,1)$ . Therefore, one can show, $(\ln \lambda)^2 = (\ln a)^2$ $\Rightarrow \ln \lambda = \pm\ln a$ $\Rightarrow \lambda = a, \frac{1}{a}$

$a$ is within the interval $[a,b]$ (inclusive), but $\frac{1}{a}$ , a valid special mean, certainly is not.