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: 2logx2^{logx} + 2logx2\sqrt2^{logx^{2}} =8

3) The sum of all integral values of a ϵ[9,9]\epsilon [-9,9] so that the equation (x2)log13(3x7a)(x-2)log_{13}(3^{x}-7a) = log132log_{13}2 + 2log13a2log_{13}a has integral solutions.

4) Find difference in the number of integer values of y and x in solution of system of inequalities: log2x(2y)>0log_{2-x}(2-y)>0 and log4y(2x2)>0log_{4-y}(2x-2)>0 .

5) Solve: 8.27log6x 8.27^{log_6x} + 27.8log6x27.8^{log_6x} - x3x>0x^{3}-x>0

Any help would be greatly appreciated.

Note by Baibhab Chakraborty
4 months, 1 week ago

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Answer to second question -

8=4+4=22+24=22+222logx=28 = 4 + 4 = 2^{2} + \sqrt{2}^{4} = 2^{2} + \sqrt{2}^{2^{2}} \Rightarrow \log x = 2 x=100 \therefore \boxed{x = 100}

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

A Former Brilliant Member - 4 months, 1 week ago

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Thank you sir.

Baibhab Chakraborty - 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)

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez 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

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member Hmm okay sir

Jason Gomez - 4 months, 1 week ago

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

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member Okay गंदा पानी

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez Dirty water? lmao

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member I wanted it to rhyme(but ofc you wouldn’t know it rhymed)

Jason Gomez - 4 months, 1 week ago

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For the first one use a1=3,a2=5,a3=32,a4=3×5,a5=52,a6=33a_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

Jason Gomez - 4 months, 1 week ago

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Or can n tend to infinity?

Jason Gomez - 4 months, 1 week ago

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

Baibhab Chakraborty - 4 months, 1 week ago

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

Baibhab Chakraborty - 4 months, 1 week ago

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

Jason Gomez - 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

Jason Gomez - 4 months, 1 week ago

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If it can the answer is just (13+132...)(15+152...)=18(\frac13+ \frac{1}{3^2} ...) (\frac15+ \frac{1}{5^2} ...)=\frac18

Jason Gomez - 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.

Baibhab Chakraborty - 4 months, 1 week ago

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@Baibhab Chakraborty Oh yeah add an one too

Jason Gomez - 4 months, 1 week ago

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

(1+13+132+133)(1+15+152+153)=1.5×1.25=1.875(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

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member Yes Baibhab had corrected me and written the same too…

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez Yeah, just saw that.

A Former Brilliant Member - 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

Jason Gomez - 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

Jason Gomez - 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.

A Former Brilliant Member - 4 months, 1 week ago

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

Jason Gomez - 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

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member But I think the log is only on one of them, else it becomes too easy to solve

Jason Gomez - 4 months, 1 week ago

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multiplication

Baibhab Chakraborty - 4 months, 1 week ago

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

A Former Brilliant Member - 4 months, 1 week ago

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Answer to fifth question, if multiplication replaces the decimals -

Let log6x=y\log_{6} x = y, then 6y=x6^{y} = x

8×27y+27×8y63y>x216(27y1+8y1216y1)>x8 \times 27^{y} + 27 \times 8^{y} - 6^{3y} > x \Rightarrow 216(27^{y-1} + 8^{y-1} - 216^{y-1}) > x

If yy is higher than 1, (27y1+8y1216y1)(27^{y-1} + 8^{y-1} - 216^{y-1}) will give us a negative value, and xx will become negative. xx cannot be negative, as 6y=x6^{y} = x.

So y1y \leq 1, and x<216(27y1+8y1216y1)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=6x = 6, because maximum value of y=1y = 1, and 6y=x6^{y} = x.

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

A Former Brilliant Member - 4 months, 1 week ago

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I think you mean 8×27y+27×8y63y>x8 \times 27^y + 27 \times 8^y - 6^{3y} > x

Jason Gomez - 4 months, 1 week ago

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

Jason Gomez - 4 months, 1 week ago

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

A Former Brilliant Member - 4 months, 1 week ago

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@Jason Gomez Now it is complete, I think...lmao

A Former Brilliant Member - 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)

Jason Gomez - 4 months, 1 week ago

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How did you get that? What did you input?

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member The exact same equation given in the beginning

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez Wolfram gives this when I input the inequality -

1.21644×109<x<9.533021.21644 \times 10^{-9} < x < 9.53302

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member x=0 satisfies the equation, so wolfram is wrong(not enough computational power?)

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez 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...

A Former Brilliant Member - 4 months, 1 week ago

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@Jason Gomez I probably did something wrong then...

A Former Brilliant Member - 4 months, 1 week ago

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

A Former Brilliant Member - 4 months, 1 week ago

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exactly sir. Thank you.

Baibhab Chakraborty - 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...

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member @Dr.Perseus no probs at all doc

Jason Gomez - 4 months, 1 week ago

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@Jason Gomez I shall dub thee...the brick boy...

A Former Brilliant Member - 4 months, 1 week ago

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@Baibhab Chakraborty - Are you really 16??

A Former Brilliant Member - 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.

Baibhab Chakraborty - 4 months, 1 week ago

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

A Former Brilliant Member - 4 months, 1 week ago

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@A Former Brilliant Member Yes XD

Baibhab Chakraborty - 4 months, 1 week ago

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@Baibhab Chakraborty thank you lmao

A Former Brilliant Member - 4 months, 1 week ago

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@Siddharth Chakravarty - Help, problem 5

A Former Brilliant Member - 4 months, 1 week ago

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@Siddharth Chakravarty - Do you play Animal Jam?

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Yes.

Siddharth Chakravarty - 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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@A Former Brilliant Member 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

Siddharth Chakravarty - 4 months ago

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@Siddharth Chakravarty you had a editor? At just 200 something subs? lmao

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@A Former Brilliant Member Editor=Editing software I meant. 😂

Siddharth Chakravarty - 4 months ago

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@Siddharth Chakravarty Oh lol ok

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@A Former Brilliant Member 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.

Siddharth Chakravarty - 4 months ago

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

Baibhab Chakraborty - 3 months, 4 weeks ago

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