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MATH WITH RAJAN
Here you can find many great Math Questions and interesting math question
find many questions about algebra, equation calculations, comparings• math Olympiad problem 🤩🤩🤩🤩
In the video,l often introduce some really fantastic topic and trick wich you will have not in the school 🏫
I will publish the videos twist perday, usually, Sunday and Saturday so subscribe to my CZcams channel so that you won't miss any videos 💯💯💯💯
MATH WITH RAJAN
Japanese - Math Olympiad question | Exponential problem solving
Japanese - Math Olympiad question | Exponential problem solving
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About -
Hello guys welcome to Math Olympiad question
Your Queries:-
exponents
exponent,rules of exponents
exponent rules
solving exponential equations
exponent problem
what is an exponent
solving exponents
maths problem
what are exponents in math
solving negative exponents
exponents rules
laws of exponents
solving exponential equations using logarithms
what are exponents
negative exponents
square root
square roots
square root problem
how to solve square root problem
math square root problem
simplify square roots
simplifying square roots
math square root problem solving
basic square root problems
simplify square root
how to simplify the square root
square root example problems
simplify square root expressions
square
how to simplify the square root of a number
root
a cambridge nested square root interview problem
math square root
solving exponential equations with different bases
practice problems
exponents and powers
exponent videos
math olympiad question
math olympiad questions
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3(81^x) = 6 81^x = (6/3) = 2 log 2 / log 81 = 0.157732438 81^ 0.157732438 = 1.999999997 1.999999997 x 3 = 5.999999999
Before watching: This will go most easily if we square away the radicals on the left. Recall that a^(mn) = (a^m)^n. First, square the left and the right, to get 3sqrt(9 sqrt(27)) = 81^(6x) = 3^(24x) (because 3^4=81) Squaring again we get (3^2)(9)sqrt(27) = 81 sqrt(27)= 81^(12x) And with one last squaring... (81^2)(27) = 81^(24x). Since 81 = 3^4, 27 = 3^3, and (a^m)(a^n) = a^(m+n)... (3^4)^2 * 3^3 = (3^4)^(24x) -> 3^11 = 3^(96x). Then 11 = 96x, and finally x = 11/96
Go away
C
7/16 in mind
Rather than squaring expressions I converted them to exponential form right away: On the left 3^1/2 x 9^1/4 x 27^1/8, all powers of 3; on the right 3^12x. Add exponents on the left to get 11/8 with 12x on the right. X = 11/96.
😂😂😂😂
Nice
@@rajanmaths5 aur kya haal hai bro??
Substitute a=b+1/2. (b+1/2)^4-(b-1/2)^4=2[4b^3(1/2)+4b(1/8)]=0 b^3+b/4=0. b=0 or b=±i/2. a=1/2 or (1±i)/2
the right answer is 262143
what a waste of time
Mate, you are among the 99% who got it wrong.
0:10 111111111111111111 but in base 2
I don't get it. Why not say 262144 - 1 ? If you are going to take the step that 512 is 2 to the ninth, why not say that 2 to the eighteenth is 262144 ?
This is super-stupid. If you know that 2^9 is 512, you could just do 512^2 - 1...factoring it into (512+1)(512-1) is the single dumbest way you could go about solving from there. But then you do a whole bunch of other time-wasting after.
...and end up with the wrong answer 🤔
oxymoron
Just do multiplication by 2 manually lol
2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144 (2^18) So 262143.
This won't work for anybody that can't multiply by 2.
@@proffessorclueless Just multiply 1024 x 256 -1. Simpler than all rearrangement. I know power of 2 up to 20 by heart (plus 2^24 and 2^32) as a low level computer guy.
@@marcelopacheco2479 sounds like binary is the way to go here.
It really is not that difficult toto just multiply 511 x 513 which equals 262,143.
Yes😂😂😂😂
I am afraid that the final sum is wrong.
Your right it should be 262,143.
X=0
ITNA LAMBA Q KIYA? 4^8/8^4 - 8^4/4^8 can also be written as 64*64*16/64*64 - 64*64/64*64*16 sare 64 cut jayenge, simple bachega 16 - 1/16 = (265-1)16 =255/16
The easiest solution is to put everything in powers of 2. You then have 2^2^8 / 2^3^4 - 2^3^4 / 2^2^8 = 2^16/2^12 - 2^12/2^16 = 2^4 - 1/2^4 = 16 - 1/16 = 255/16.
45 = 3^3.465...rien de plus simple
sqrt(10 * sqrt(10 * sqrt(10))) = 100^x (10 * (10 * 10^(1/2))^(1/2))^(1/2) = 10^2x 10 * (10 * 10^(1/2))^(1/2) = 10^4x 100 * 10 * 10^(1/2)) = 10^8x 10 000 * 100 * 10 = 10^16x 10 000 000 = 10^16x 10^7 = 10^16x 7 = 16x 7/16 = x x = 7/16 = 0,4375
He wastes time on several steps. 3^x = 45 can immediately be re-written as log_3 (45) = x. We're not required to use log base 10 to eventually get to this step.
Dividing by 2 at the beginning was unnecessary.
If 99% of entrants failed this they never learned arithmetic.
2 *sqrt(x) = x. One solution is zero. For the other, divide by sqrt(x) to get 2 = sqrt(x). That means x = 4, the other solution.
I don't believe 99% would fail to find the 0 solution. You wouldn't clickbait, would you?
I literally just looked at it and knew 3⁴ + 3² = 90. You REALLY overcomplicated it
Nearby value?
Can any suggest what is value of x? In rational
Didn't watch the video but I got x=3.465
3²+3⁴=90, but answer is 9×5+9×5=45+45=90 because we have to so how much we can remember and show that why 3×3×3×3+×3×3 is 90 Which we never use in our life. I am not telling that math and science is useless be this this 💀😕 education system crush it and fuck it
3^x=45, 3^(x-2)=5, (x-2) log 3 = log 5, x=2+(log 5)/(log 3). If 99% failed this, the population had never learned about logarithms.
easy toh hai😂😂
x=4 and possibly imaginary values
No, no imaginary answers. But zero is also a solution.
@@JohnRandomness105 Yes I missed that one.
If we divide by √x, we get 2=√x i.e. x=4. But the equation has one more solution (when division is invalid): x=0. (If x is negative, then √x is imaginary. More generally, if x is complex other than a positive real number, then angle of √x in polar coordinates can't be the same as the angle of x.) We can also substitute y=√x; then we get y^2 - 2*y = 0, and that yields y*(y-2)=0, immediately giving the two solutions (a quadratic equation can't have more than two solutions).
Nice calculating. But how on earth are you writing the letter "x"?
I think they're using n, same concept.
@@FinetalPies The Thumbnail says "x", and "n" implies a natural number.
1:27 x^3 + x^2 + x - 555 = 0👀 1 1 1 -155 Syntetic division 5 5 30 155 1 6 31 0 x^2 + 6x + 31=0 D1=(b/2)2-ac=3^2-1*31 < 0
Acho que está errado essa resolução.
Isn't that not simple 7+49+343+2401+16807 = 19607
25
B 25
❤❤❤
Silly soundtrack. Left after 20 seconds
Oh wow thank you
Hahaha
Isn't that just a simple sum of gp?
The final result would still have 7 raised to 5
Muito legal!
5^2/25^2 = (5/25)^2 = (1/5)^2 = 1^2/5^2 =1/25