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I came across this question in ISI(Indian Statistical Institute) admission test

$$I=\int_2^3 \frac{dx}{\ln(x)} $$

The four options were

(A) is less than $2$

(B) is equal to $2$

(C) lies in the interval $(2, 3)$

(D) is greater than $3$

How do I go about solving this problem using high school level calculus?

Any help is greatly appreciated!

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3 Answers 3

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We have

$$2\ln\left(\frac{\ln 3}{\ln 2}\right)=2\int_2^3\frac{dx}{x\ln x}\le\int_2^3\frac{dx}{\ln x}\le 3\int_2^3\frac{dx}{x\ln x}=3\ln\left(\frac{\ln 3}{\ln 2}\right)$$ Can you now choose the right option?

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  • $\begingroup$ I don't think this is how they wanted them to proceed: making use of this requires you to figure out that $\ln(\ln(3)/\ln(2))$ is less than $2/3$, which seems a bit tedious to prove. (In particular, you have to do better than the crude approximation $\ln(3)/\ln(2)=\log_2(3)<\log_2(4)=2$ to get this result, since $\ln(2)>2/3$.) $\endgroup$
    – Ian
    Mar 17, 2016 at 18:03
  • $\begingroup$ We have $$\ln(3^3)=3\ln 3<\ln(2^5)=5\ln 2$$ so $$\frac{\ln3}{\ln2}<\frac53$$ and with the inequality $\ln(1+x)\le x$ we get $$\ln\left(\frac{\ln 3}{\ln2}\right)\le \frac23$$ and I can answer as I want not as they want ;-) $\endgroup$
    – user296113
    Mar 17, 2016 at 18:12
  • $\begingroup$ That's pretty slick (it would have taken me a while to come up with the trick of bringing the exponent back in), but it just barely does the job, whereas the simple pointwise upper bound easily does the job (knowing only $\ln(2)>1/2$). $\endgroup$
    – Ian
    Mar 17, 2016 at 18:16
  • $\begingroup$ I gave a proof without using the approximation $\frac1{\ln 2}\sim 1.44$ which need a calculator. If you accept the use of calculator you should accept the use of Wolfram alpha. $\endgroup$
    – user296113
    Mar 17, 2016 at 18:21
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    $\begingroup$ As I said, the result from the other answer can be achieved only knowing $\ln(2)>1/2$, which is readily seen to be equivalent to $e<4$. I think a student at this level could take $e<4$ as a given fact. (Or, if not, they can use series techniques to conclude $\ln(2)>1-1/2=1/2$.) $\endgroup$
    – Ian
    Mar 17, 2016 at 18:25
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HINT: What is the smallest/largest $\dfrac1{\ln x}$ can be when $2\le x\le 3$?

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  • $\begingroup$ Hint: using the fact that $e\simeq2.7$ which is in the interval of interest. $\endgroup$
    – user164550
    Mar 17, 2016 at 17:46
  • $\begingroup$ So like maxima and minima? I find the derivative? I apologise, I'm not too good with this.. :/ $\endgroup$
    – Nikhil
    Mar 17, 2016 at 17:46
  • $\begingroup$ Just think about the graph of $\ln x$ and the graph of $1/(\ln x)$. $\endgroup$
    – Ted Shifrin
    Mar 17, 2016 at 17:49
  • $\begingroup$ Ah, okay! So by the looks of it from desmos by adding up the areas from 3 to 2 for both of them, it looks like it'll be less than 2, but is there any other way to do this? $\endgroup$
    – Nikhil
    Mar 17, 2016 at 17:53
  • $\begingroup$ The biggest possible area would be the biggest height of the function ($1/(\ln 2) \approx 1.44$) times the length of the interval ($1$). $\endgroup$
    – Ted Shifrin
    Mar 17, 2016 at 17:59
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Graphically :

Let $I=\int_2^3 \frac{1}{ln(x)}dx$

The following graph is obviously not to scale, but is useful to get yourself an idea of what we're doing.

The area of the blue rectangle* is $A_r=\frac{1}{ln(3)}$

The area of the red triangle is $A_t=\frac{1}{2} \times 1 \times (\frac{1}{ln(2)}-\frac{1}{ln(3)})$

We have $I \leqslant A_r+A_t=\frac{1}{2} \times (\frac{1}{ln(2)}+\frac{1}{ln(3)}) = \frac{ln(2)+ln(3)}{2ln(2)ln(3)} \leqslant \frac{1}{ln(2)} < 2$ since $\sqrt e <\sqrt3<1.8 < 2$

The first inequality comes from the fact that your function is convex (you can check the second derivative is positive on $[2,3]$)

edit : yeah, come to think of it, it was a bit silly, you could just find $I\leqslant \frac{1}{ln(2)}$ immediately as other answers stated already... still $\frac{1}{2} (\frac{1}{ln(2)}+\frac{1}{ln(3)})$ is a good upper bound so I guess I'll still leave it anyway.

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