Problem Statement:-

Find the set of real values of $x$ for which $$x^{(\log_{10}x)^2-3\log_{10}x+1}\gt1000$$

Correct Approach:-

Domain of $x^{(\log_{10}x)^2-3\log_{10}x+1}\implies x\gt0$

Now, $$x^{(\log_{10}x)^2-3\log_{10}x+1}\gt1000$$

Taking $\log_{10}$ on both sides, we get


Let $z=\log_{10}{x}$

Hence, $(z^2-3z+1)z\gt3\implies z^3-3z^2+z-3\gt0$

Fortunately for factorising the polynomial we dont have to go too far we get the first factor at $z=3$. So, on factorising, we get $$(z-3)(z^2+1)\gt0$$

As, $\forall z\in\Bbb{R}.(z^2+1\gt0)$, hence $$z-3\gt0\implies z\gt3\implies \log_{10}x\gt3\implies x\gt1000$$

Incorrect Approach:-

Domain of $x^{(\log_{10}x)^2-3\log_{10}x+1}\implies x\gt0$

We also see that for $x=1$ the inequality is not satisfied, so we would not be eliminating any values from the interval satisfying the inequality by putting the condition $x\neq1$

Now, take $\log_{x}$ on both sides to get

$$((\log_{10}x)^2-3\log_{10}x+1)\gt\dfrac{3}{\log_{10}x}\\ \implies ((\log_{10}x)^2+1)\gt\dfrac{3}{\log_{10}x}+3\log_{10}x$$

Using $A.M.\ge G.M.$, we get $$((\log_{10}x)^2+1)\gt6\implies (\log_{10}x)^2\gt5\implies -\sqrt{5}\lt\log_{10}x\lt\sqrt5\\ \implies 10^{-\sqrt{5}}\lt x\lt10^{\sqrt5}$$

But, as we had excluded $1$ as a solution, so the interval satisfying the inequality comes out to be $x\in(10^{-\sqrt5},10^{\sqrt5})-\{1\}$

I have tried two different approaches to the problem, while one does provide the correct answer and the other doesn't but what is wrong with the incorrect approach.


There are a couple of mistakes in your last approach

In using $AM \ge GM$ you assume that $z > 3$, which I guess you got from the first result. What follows from there is

$$ \frac{z^2 + 1}{2} > z > 3 \quad\Rightarrow\quad z^2 > 5 $$

What you can conclude from this is

$$ |z| > \sqrt{5} \quad\Rightarrow\quad z > \sqrt{5} \quad\mbox{or}\quad z < -\sqrt{5} $$

The second region is in contradiction with what you assumed before ($z>3$), so you end up with $z > \sqrt{5}$. So the region is $\{z : z > \sqrt{5}\mbox{ and } z > 3\} = \{z : z > 3\}$ which is the first result

  • $\begingroup$ If you assume $z=\log_{10}x$ as in my correct approach, then I didn't assume anything for $z$ in my incorrect approach. In fact, I had tried the incorrect approach first rather than the correct one. Also it seems in your statement-"In using $AM \ge GM$ you assume that $z > 3$, which I guess you got from the first result." you are telling that it is necessary to assume $z\gt3$ to apply $AM \ge GM$ $\endgroup$ – user350331 Nov 27 '16 at 2:49
  • $\begingroup$ @user350331 Perhaps I misunderstood, but how do you conclude that $z^2 + 1 > 6$? $\endgroup$ – caverac Nov 27 '16 at 2:56
  • $\begingroup$ By using $AM\ge GM$ on $\dfrac{3}{z}+z$ to get $\dfrac{3}{z}+z\ge2\times3\implies \dfrac{3}{z}+z\ge6$, I might just add a little something which makes a lot of difference and I might just start to see why my second answer is faulty, which is $z\gt0\implies x\gt1$. And, similarly for $z\lt0\implies x\lt1$, we get $\dfrac{3}{z}+z\le-6$, this case is trivial to consider since we would arrive at $z^2+1\gt-6$ which is trivially true. $\endgroup$ – user350331 Nov 27 '16 at 3:07


Using $A.M.\ge G.M.$, we get $$((\log_{10}x)^2+1)\gt6$$

One problem is that you can't use AM-GM to begin with, since at this point it hasn't been established that $\log_{10} x \ge 0$ and AM-GM only works for non-negative numbers.

The major problem with this step, however, is that you essentially need to prove $a \gt b$, but instead you use $a \gt b \gt c$ and prove $a \gt c$, yet this does not prove the original inequality $a > b$.

Consider for example how the same (wrong) logic could be used to prove the obviously false: $$4+4 \gt 1+8$$

Using $A.M.\ge G.M.$, we get

$$4+4 \gt 2 \sqrt{8} = 4 \sqrt{2}$$

which is true, even though the inequality we needed to prove $8 \gt 9$ is false.


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