# 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
1 month ago

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

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

- 1 month ago

Thank you sir.

- 1 month ago

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)

- 1 month ago

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

- 1 month ago

Hmm okay sir

- 1 month ago

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

- 1 month ago

Okay गंदा पानी

- 1 month ago

Dirty water? lmao

- 1 month ago

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

- 1 month ago

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

- 1 month ago

Or can n tend to infinity?

- 1 month ago

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

- 1 month ago

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

- 1 month ago

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

- 1 month ago

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

- 1 month ago

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

- 1 month ago

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.

- 1 month ago

Oh yeah add an one too

- 1 month ago

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

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

- 1 month ago

Yes Baibhab had corrected me and written the same too…

- 1 month ago

Yeah, just saw that.

- 1 month ago

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

- 1 month ago

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

- 1 month ago

@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.

- 1 month ago

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

- 1 month ago

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

- 1 month ago

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

- 1 month ago

multiplication

- 1 month ago

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

- 1 month ago

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?

- 1 month ago

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

- 1 month ago

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

- 1 month ago

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

- 1 month ago

Now it is complete, I think...lmao

- 1 month ago

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

- 1 month ago

How did you get that? What did you input?

- 1 month ago

The exact same equation given in the beginning

- 1 month ago

Wolfram gives this when I input the inequality -

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

- 1 month ago

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

- 1 month ago

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...

- 1 month ago

I probably did something wrong then...

- 1 month ago

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

- 1 month ago

exactly sir. Thank you.

- 1 month ago

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

- 1 month ago

@Dr.Perseus no probs at all doc

- 1 month ago

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

- 1 month ago

@Baibhab Chakraborty - Are you really 16??

- 1 month ago

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

- 1 month ago

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

- 1 month ago

Yes XD

- 1 month ago

thank you lmao

- 1 month ago

@Siddharth Chakravarty - Help, problem 5

- 1 month ago

@Siddharth Chakravarty - Do you play Animal Jam?

- 3 weeks, 4 days ago

Yes.

- 3 weeks, 4 days ago

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

- 3 weeks, 3 days ago

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

- 3 weeks, 3 days ago

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

- 3 weeks, 3 days ago

Editor=Editing software I meant. 😂

- 3 weeks, 3 days ago

Oh lol ok

- 3 weeks, 3 days ago

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.

- 3 weeks, 3 days ago

lol okay

- 3 weeks, 3 days ago

Lol

- 3 weeks, 3 days ago

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

- 3 weeks ago