Exercise 1.13 [Section 1, BOMC-2011]

[sourav das]

@Joti. Don't be afraid of IMO problems. If you are a student of class 9 and read Higher Mathematics text book, I think you have already met with an IMO standard problem. If you don't, try the following one:
IMO Longlisted Problem 1978 ( Proposed by Turkey)
Simplify\[\frac{1}{\log_{a}abc }+\frac{1}{\log_{b}abc}+\frac{1}{\log_{c}abc}\]
$a,b,c>0$

And your solution doesn't seem to be wrong except the $p^2$ part.

[Labib]

What did he eat before setting this prob, huh?? Turkey??

[sm.joty]

Thanks to Sourav da.

What did he eat before setting this prob, huh?? Turkey??

@Labib hi....hi....hi

[Jini]

This is how I did it:

$\frac{(4x^2)}{(1-\sqrt{1+2x})^2}$
$= \frac{(4x^2)}{(1-\sqrt{1+2x})^2} . \frac{(1+\sqrt{1+2x})^2}{(1+\sqrt{1+2x})^2}$
$= \frac{(4x^2)(1+\sqrt{1+2x})^2}{(1^2-(1+2x))^2}$
$= \frac{(4x^2)(1+\sqrt{1+2x})^2}{(-2x)^2}$
$= (1+\sqrt{1+2x})^2$

Now, we are given:

$\frac{(4x^2)}{(1-\sqrt{1+2x})^2}<2x+9$
therefore,
or $(1+\sqrt{1+2x})^2<2x+9$
or $1+2\sqrt{1+2x}+\sqrt{1+2x}^2<2x+1+8$
or $2(1+2x)^\frac{1}{2}<7$
or $1+2x<\frac{49}{4}$
or $x< \frac{45}{8}$

I actually found this quite easy... Did I do something wrong? :S

This is actually the first time I'm posting a here, so forgive me if how I wrote what I wrote is hard to read (it was exhausting to write!)

Jini

[Jini]

@Sourav:

$\frac{1}{log_{a}abc}+\frac{1}{log_{b}abc}+\frac{1}{log_{c}abc}$

$a;b;c>0$

I wonder if this is right:

$\frac{1}{log_{a}abc}+\frac{1}{log_{b}abc}+\frac{1}{log_{c}abc}$
$= log_{abc}a + log_{abc}b + log_{abc}c$
$= log_{abc}abc$
$= 1$

[sourav das]

@Jini you have included the case when $x=0$ and miss the part $x \geq - \frac{1}{2}$
And for the IMO longlisted problem:
(i) If you're a student under class -9 "Well done"
(ii)If you're a student of upper class-9 then....

[*Mahi*]

Both of the problems uses tricks which are taught in our high school, but that doesn't mean that they are negligible. On the other hand , it is a good thing that you learned and can use those right!

[Jini]

@Jini you have included the case when $x=0$ and $x \geq - \frac{1}{2}$
And for the IMO longlisted problem:
(i) If you're a student under class -9 "Well done"
(ii)If you're a student of upper class-9 then....

The answer comes out to be imaginary otherwise... Should I add the lower limit, then?

About the other thing, I haven't done logarithm for a really long time... I thought a quick check if I remember would be nice... Imagine if I'd got it wrong!

[*Mahi*]

As he wrote, the lower limit $\frac 1 2$ , and $x \not = 0$ should be mentioned in the solution.

[Jini]

You mean $ -\frac{1}{2}$? Doesn't $x=0$ fall between the two limits anyway?