If $x, \log_{10}(x) , \log_{10}\log_{10}(x)$ are in arithmetic progression, find the range of $x.$

(a) $0 < x < 1$

(b) $1 < x < 10$

(c) $10 < x < 100$

(d) $100 < x < 1000$

I have found the answer but I want a solution using logic without graph.

  • $\begingroup$ I re-formatted your question, please double check I didn't change anything. $\endgroup$ – user2468 Aug 19 '12 at 17:42
  • $\begingroup$ What is the base of the log ? $\endgroup$ – Belgi Aug 19 '12 at 17:43
  • $\begingroup$ @Belgi $\log_{10} $ $\endgroup$ – Pedro Tamaroff Aug 19 '12 at 17:43
  • 2
    $\begingroup$ In case anyone is wondering, the progression turns out to be approximately $1.22802, 0.0892054, -1.04961$. $\endgroup$ – Rahul Aug 19 '12 at 18:16

To say that these numbers are in arithmetic progression is to say that $$ 2 \, \log_{10} x = x + \log_{10} \log_{10} x \, . $$ Exponentiating this gives the equivalent equation $$ x^2=10^x\log_{10} x \, . $$ If $x<1$, the two sides of this equation have opposite sign, so the equation doesn't hold. Moreover, if $x > 10$, $$ 10^x \log_{10} x > 10^x = (\sqrt{10})^{2x} > (2^{x})^2 > x^2 \, , $$ so again the equation doesn't hold. So the only way it can hold is if $1<x<10$.

On the other hand, the function $f(x)=x^2-10^x \log_{10} x$ is continuous and has a sign change on $[1,10]$, so by the Intermediate Value Theorem there must be some $x$ in that range which satisfies the equation.


I have found one approach If If $\log a$, $\log b$ and $\log c$ are in AP then $a$, $b$ and $c$ are in GP.

Hence, $x^{2}=10^{x}\times \log_{10}x$.

Among the given options this is only possible if x is less than 10 and greater than 1.


If $x$ is less than $1$, then $\log x$ is negative, and $\log \log x$ is undefined. So (a) is out. Let's look at (b). If $1 < x < 10$, then $0<\log x<1$, and $\log\log x < 0$. Maybe.

Now (c) and (d). If $10<x < 1000$, then $1<\log x <3$, and $0<\log\log x<\log 3 < 0.5$. We see that the distance between $\log x$ and $\log\log x$ is smaller than the distance between $x$ and $\log x$.

So (b) must be the answer.

We can show that there exists an $x$ in that range that makes $x$, $\log x$, $\log \log x$ an arithmetic progression. As $x$ moves from $1$ to $10$,

  • the difference between $x$ and $\log x$ moves continuously from $1-0$ to $10 - 1$, that is, from $1$ to $9$.
  • the difference between $\log x$ and $\log \log x$ moves continuously from infinity to $1-0 = 1$.

Therefore we must have $x - \log x = \log x - \log \log x$ for some $x$ in the interval $(1,10)$.


I know this is tagged "precalculus", but the following comes to mind ...

Note that if $f(x)=\log_{10}x \text{ then } f'(x)=\cfrac 1 {x \log_e {10}}$

Apply the Mean Value Theorem to $f(x)=\log_{10}x$ at the points $x$ and $y$ with $x>y$ to identify a $z$ with $y < z < x$ such that $\log_{10}x-\log_{10}y = \cfrac 1 {z \log_e {10}}(x-y)$.

Now set $y=\log_{10}x$ and use the arithmetic progression to see that the differences are equal so that $\cfrac 1 {z \log_e {10}}=1 \text { and } z=\cfrac 1 {\log_e {10}} \approx 0.4$.

So we have $y<z$ so $10^y (=x) < 10^z (= e$). And it is easy to see that $x>1$ (for $\log \log x$ to exist).

So to conclude, we have $1 < x < e$.


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