# Doubts.

1) If $$a_1,a_2,a_3,a_4,........a_n$$ are distinct odd natural numbers such that they aren't divisible by no prime number greater than 5, then maximum value of $$\sum_{i=1}^n \frac{1}{a_i}$$ is

2) Solve: $2^{logx}$ + $\sqrt2^{logx^{2}}$ =8

3) The sum of all integral values of a $\epsilon [-9,9]$ so that the equation $(x-2)log_{13}(3^{x}-7a)$ = $log_{13}2$ + $2log_{13}a$ has integral solutions.

4) Find difference in the number of integer values of y and x in solution of system of inequalities: $log_{2-x}(2-y)>0$ and $log_{4-y}(2x-2)>0$ .

5) Solve: $8.27^{log_6x}$ + $27.8^{log_6x}$ - $x^{3}-x>0$

Any help would be greatly appreciated.

Note by Baibhab Chakraborty
4 months, 1 week ago

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## Comments

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Answer to second question -

$8 = 4 + 4 = 2^{2} + \sqrt{2}^{4} = 2^{2} + \sqrt{2}^{2^{2}} \Rightarrow \log x = 2$ $\therefore \boxed{x = 100}$

I will try to solve the rest and send later @Baibhab Chakraborty :)

- 4 months, 1 week ago

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Thank you sir.

- 4 months, 1 week ago

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He is of your age too, I don't think he will like to be called sir either( but if Percy wants to be addressed so, then so be it)

- 4 months, 1 week ago

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No please don't call me sir, Percy is just fine @Baibhab Chakraborty

@Jason Gomez - I do not want to be addressed like that...don't give people ideas lmao

- 4 months, 1 week ago

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Hmm okay sir

- 4 months, 1 week ago

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Don't call me sir, दादी (google translate copy/paste lmao)

- 4 months, 1 week ago

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Okay गंदा पानी

- 4 months, 1 week ago

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Dirty water? lmao

- 4 months, 1 week ago

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I wanted it to rhyme(but ofc you wouldn’t know it rhymed)

- 4 months, 1 week ago

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For the first one use $a_1=3,a_2=5,a_3=3^2,a_4=3 \times 5,a_5=5^2, a_6 = 3^3$ till where required

- 4 months, 1 week ago

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Or can n tend to infinity?

- 4 months, 1 week ago

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n is a variable, I guess it might not need the value of n whatsoever.

- 4 months, 1 week ago

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Sir, can you please explain why? Thank you sir for paying heed to my doubt.

- 4 months, 1 week ago

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Please don't call me sir, I am just 17 years old :)

- 4 months, 1 week ago

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You don't want the numbers to be even, so no factors of two are allowed, at the same time you want the only prime factors to be less than or equal to 5, so just take the prime factors as 3 and 5 and their powers and multiples as the numbers

- 4 months, 1 week ago

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If it can the answer is just $(\frac13+ \frac{1}{3^2} ...) (\frac15+ \frac{1}{5^2} ...)=\frac18$

- 4 months, 1 week ago

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Shouldnt it start from 1 because 1 makes the sum greater as then 3/2.5/4=15/8 . This might be the maximum value.

- 4 months, 1 week ago

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Oh yeah add an one too

- 4 months, 1 week ago

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Yes, its just the product of two Geometric series' I think.

$(1 + \dfrac{1}{3} + \dfrac{1}{3^{2}} + \dfrac{1}{3^{3}} \ldots)(1 + \dfrac{1}{5} + \dfrac{1}{5^{2}} + \dfrac{1}{5^{3}} \ldots) = 1.5 \times 1.25 = 1.875$

- 4 months, 1 week ago

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Yes Baibhab had corrected me and written the same too…

- 4 months, 1 week ago

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Yeah, just saw that.

- 4 months, 1 week ago

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For the fourth one, the both inequalities have infinite solutions, so you can’t compare both

- 4 months, 1 week ago

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I don’t know how to solve the third one, but “a” has to be greater than zero for the right hand side to be well defined

- 4 months, 1 week ago

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@Baibhab Chakraborty - Is the fifth one 8 x 27 or 8.27 because if it is multiplication instead of decimal, I think I can solve it.

- 4 months, 1 week ago

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I think it’s multiplication(decimal will be just too bad)

- 4 months, 1 week ago

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Yea, it will be very messy. 8 x 27 should work well because the power log is at base 6, and 8 x 27 is just 6^3

- 4 months, 1 week ago

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But I think the log is only on one of them, else it becomes too easy to solve

- 4 months, 1 week ago

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multiplication

- 4 months, 1 week ago

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Okay. Use \times in your latex to get the multiplication cross symbol :)

- 4 months, 1 week ago

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Answer to fifth question, if multiplication replaces the decimals -

Let $\log_{6} x = y$, then $6^{y} = x$

$8 \times 27^{y} + 27 \times 8^{y} - 6^{3y} > x \Rightarrow 216(27^{y-1} + 8^{y-1} - 216^{y-1}) > x$

If $y$ is higher than 1, $(27^{y-1} + 8^{y-1} - 216^{y-1})$ will give us a negative value, and $x$ will become negative. $x$ cannot be negative, as $6^{y} = x$.

So $y \leq 1$, and $x < 216(27^{y-1} + 8^{y-1} - 216^{y-1})$

I don't know what exactly the question is looking for, but my best guess is a maximum value, since we have lesser than and lesser than equal to, so in that case, maximum value of $x = 6$, because maximum value of $y = 1$, and $6^{y} = x$.

@Baibhab Chakraborty - Is this what you were looking for?

- 4 months, 1 week ago

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I think you mean $8 \times 27^y + 27 \times 8^y - 6^{3y} > x$

- 4 months, 1 week ago

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You got every term on the left hand side wrong, congrats!!

- 4 months, 1 week ago

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Bruh, I was still editing, just went AFk for a few moments

- 4 months, 1 week ago

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Now it is complete, I think...lmao

- 4 months, 1 week ago

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According to desmos the value of x lies between 0 and 9.533 (all values in between satisfy)

- 4 months, 1 week ago

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How did you get that? What did you input?

- 4 months, 1 week ago

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The exact same equation given in the beginning

- 4 months, 1 week ago

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Wolfram gives this when I input the inequality -

$1.21644 \times 10^{-9} < x < 9.53302$

- 4 months, 1 week ago

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x=0 satisfies the equation, so wolfram is wrong(not enough computational power?)

- 4 months, 1 week ago

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If you use x=0 in the equation, then log x just approaches negative infinity, right? So the lhs would then just approach zero? I'm confused...

- 4 months, 1 week ago

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I probably did something wrong then...

- 4 months, 1 week ago

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nvm, I input the 3 inequalities and got 0 to 9.533 as well

- 4 months, 1 week ago

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exactly sir. Thank you.

- 4 months, 1 week ago

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Uh...can you, like not call me sir? Its embarrassing, I'm the same age as you...

- 4 months, 1 week ago

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@Dr.Perseus no probs at all doc

- 4 months, 1 week ago

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I shall dub thee...the brick boy...

- 4 months, 1 week ago

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@Baibhab Chakraborty - Are you really 16??

- 4 months, 1 week ago

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Ah yes, I turned 16 this January only. In fact, I joined Brilliant this February. Thank you sir for your kind help.

- 4 months, 1 week ago

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Cool, I'm 16 too, so can you please stop calling me sir?

- 4 months, 1 week ago

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Yes XD

- 4 months, 1 week ago

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thank you lmao

- 4 months, 1 week ago

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@Siddharth Chakravarty - Help, problem 5

- 4 months, 1 week ago

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@Siddharth Chakravarty - Do you play Animal Jam?

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Yes.

- 4 months ago

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Okay. So there's this Animal Jam Tuber named Siddharth Chakravarty, who also happens to have commented on Mahdi's yt video. Coincidence, I think not. If you really are that person, all I can say is, your voice doesn't sound right for a 16 yr old...lmao

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Lol that's me only but the videos are too old, liks 2 years ago and also my editor used to make mine voice a bit effiminate. XD

- 4 months ago

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you had a editor? At just 200 something subs? lmao

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Editor=Editing software I meant. 😂

- 4 months ago

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Oh lol ok

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Lol I just watched my videos with the voice and it felt so cringy 😂 BTW count me out of Mathathon as I just remembered about it today suddenly and saw now that you posted a lot of problems.

- 4 months ago

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lol okay

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Lol

- 4 months ago

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@Percy Jackson @Siddharth Chakravarty-can you please check my feed and help me on my latest doubt....

- 3 months, 4 weeks ago

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