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Dennis Online Math Academy
Kenya
เข้าร่วมเมื่อ 28 พ.ย. 2021
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Math Olympiad| A Nice Math Olympiad Algebra Problem.
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Harvard University Entrance Question
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I trust you are all doing fine🙏.You are welcome to solve this Nice Math Olympiad Algebra equation .Kindly like, share, comment and Subscribe
If you like this video on how to solve this nice Math Problem, like and Subscribe to my channel for more exciting videos
Mathematics Olympiad,Nice Math Olympiad Exponential problem,Harvard University Admission Tricks, Exponential equation
Oxford University Admission exam interview tricks
Harvard University Entrance Question
Math Olympiad, Algebra, Simplification, Germany Math Olympiad,Math problem-solving, Grade 8 Maths, Grade 9 Maths, Nice Square Root Simplification, Japanese Can you solve, Olympiad High school Competition,Advanced math concepts,Math competition,KCSE mathematical equations, Find the Value of M,Math Olympiad strategies, Olympiad Problem,
Calculus,Algebraic expressions, Olympiad mathematics
#matholympiad #algebra#maths#Exponential equation #simplification #Exponents#howto#mathematics#viral #mathematicslesson#calculus # Dennis online Math Academy
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Math Olympiad X^3+X=68| A Tricky question from Oxford University
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Hello my TH-cam family I trust you are all doing fine🙏.You are welcome to solve for the value of X in this Nice Math Olympiad Algebra problem .Kindly like, share, comment and Subscribe If you like this video on how to solve this nice Math Problem, like and Subscribe to my channel for more exciting videos Mathematics Olympiad,Nice Math Olympiad Exponential problem,Harvard University Admission Tr...
Japanese Olympiad mathematics|Math Olympiad Exponential Problem.
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Hello my TH-cam family I trust you are all doing fine🙏.You are welcome to solve for the value of X in this Nice Math Olympiad exponential equation .Kindly like, share, comment and Subscribe If you like this video on how to solve this nice Math Problem, like and Subscribe to my channel for more exciting videos Mathematics Olympiad,Nice Math Olympiad Exponential problem,Harvard University Admissi...
Germany Olympiad mathematics| Exponential Equation
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A Nice Math Olympiad Exponential Equation| 7^x-2=91
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Germany|Can you Solve this?| A Nice Math Olympiad Square Root Problem.
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Germany Olympiad Mathematics|Math Olympiad Exponential Equation
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Harvard University Admission interview Tricks| Find the value of Y=?
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Germany | Can you Solve this?||A Nice Math Olympiad Algebra Problem.
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Hello my TH-cam family I trust you are all doing fine🙏.You are welcome to solve for the value of K in this Nice Math Olympiad Algebra problem .Kindly like, share, comment and Subscribe If you like this video on how to solve this nice Math Olympiad Problem, like and Subscribe to my channel for more exciting videos Mathematics Olympiad,Nice Math Olympiad Exponential problem,Harvard University Adm...
Germany| Can you Solve this?| Math Olympiad Algebra Problem.
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Germany| Can you Solve this| Math Olympiad Algebra Problem
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Hello my TH-cam family I trust you are all doing fine🙏.You are welcome to solve for the value of X in this Nice Math Olympiad Algebra problem .Kindly like, share, comment and Subscribe If you like this video on how to solve this nice Math Olympiad Problem, like and Subscribe to my channel for more exciting videos Mathematics Olympiad,Nice Math Olympiad Exponential problem,Harvard University Adm...
Oxford University Admission Entrance interview Tricks.
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Germany| Can you Solve this?| A Nice Math Olympiad Algebra Problem.
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A Nice Math Olympiad Exponential Problem.
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Germany|Can you Solve this?| A Nice Math Olympiad Algebra Problem| X=?
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Hello my TH-cam family I trust you are all doing fine🙏.You are welcome to solve for the value of X in this Nice Math Olympiad Algebra Problem .Kindly like, share, comment and Subscribe If you like this video on how to solve this nice Math Problem, like and Subscribe to my channel for more exciting videos Mathematics Olympiad,Nice Math Olympiad Exponential problem,Harvard University Admission Tr...
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Germany| Can you solve this?| A Nice Math Olympiad Algebra Problem ( x,y)= ?
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Germany| Can you solve this?|A Nice Math h Olympiad Algebra Problem
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Germany| Can you solve this?|A Nice Math h Olympiad Algebra Problem
by faktoring , (x-9)(x^2+8x+72)=0 , x=9 , x=(-8+/-V(64-288))/2 , x= -4+i*V56 , -4-i*V56 ,
@@prollysine Nice Alternative 👍
Its doubled squared in the photo
I like to add a way to evaluate 3^9 3^9=3^4*3^4*3 =81*81*3 =6561*3=19683 8 1 * 8 1 --------------------------- 64 1 1 6 ---------------------------- 6 5 61 Or 81*81=81^2=(80+1)^2 =6400 +1+2 80*1 =6401+160=6561
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Thank you.
I just looked at the thumbnail, thought about it for like 30 seconds and got the answer
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It's blatantly obvious if you've worked in I.T. or are used to thinking in binary, and 4- or 8-bit bytes!
This was too easy
There are probably much more complicated ways to solve this problem 🤣 1. X != 0, 2. 5*5 = 25 3. x*x=x2 4. multiply the equation with 5 and divide by x (because it has to be != 0, still an assumption) 5. 5³ / x³ = 1 ==> x³ = 5³ ==> x = 5, because 5³ > 0 . That's it, right?
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7^x + 7^x =490 > 7^x =245 x =log 245 (base 7) = log (7^2*5)(base 7) = log 7^2(base 7) + log 5(base7) = 2log 7(base 7) + log 5(base 7) = 2*1 + log 5(base7) = 2 + log 5(base 7)
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In the last line you wrote 2+ log 5/log 7 And in margin wrote log a /log b = log a (base b) But in answer you did not write 2+ log 5(base 7)
@prithwiraj this is clearly explained. Thanks
This is clearly explained as 7^2+log 5( base 7)
EXPONENCIAL
Simpler solution: Sqr root both side gives (x+3)^2=+/-2^2 Sqr root again gives (x+3)= +/-2 or +/-2i x = -1 or -5 or -3+2i or -3-2i 🤷🏻
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At 2:30 you make and error saying that it = a^2-ab+b^2. It is, in fact = A^2-2ab-b^2. You use the correct expansion in the square brackets after that though.
Thanks
Damn bruh
My way: 3ˣ·3ˣ·3ˣ = 30 => 27ˣ = 30 => x = ln(30)/ln(27) or x = log₃(30)/3 🙂
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7^x+7^x 7o^70= 70*70=70^2= 7^20]=[ 2*7^x =7^2x= 7^2o 2x= 2ox=1o ACADEMIC Marcelius Martirosianas
Use Euler's equation, x_j=2*cos(jπ/2)+2*i*sin(jπ/2)-3, j=0,1,2,3, to obtain four roots, where i=√(-1)
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3
That is much too circumstantial. You should cancel as much as possible in the very beginning: 7^x + 7^x = 490 => 2*7^2 = 2*49*5 => 7^x = 7^2 * 5 =>7^(x-2) = 5 => ln(7^(x-2)) = ln(5) => (x-2)*ln(7) = ln(5) => (x-2) = ln(5)/ln(7) => x = ln(5)/ln(7) + 2. Done.
Thank you well explained ❤
Much welcome.
It is obvious from the very first view when you look at it that x=5. 5 over 5 is 1. 1 multiplied by 1 is 1.
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05:34 you wrote [b to_the_power log(a/b) = b] but obviously you wanted to write [b to_the_power log(a/b) = a] because this is what you substitute a bit later, correct?
That is perfectly correct. Thank you for the comment
@slacroixfr OK to use the caret ( ^ ) for exponents: 2^3 = 8 c^2 = c*c
@@EdLeeSB I thought of it but thought some might not understand...
@@dennismathacademy side note : I was actually wrong to write "log(a/b)" that means literally log of (a divided by b) although you meant logBase(b) of a Now, I do not know how to better write "logBase(b) of a" in simple text mode (with one single font size). In some programming languages we can use log(a,b) with log( expression [ , base ] ) "[ , base]" meaning "optional with default value of e"
X=log96/log8 X=2.194875
Or this way to get the solution (step by step) ... 7^(x + 1) + 7^(x - 1) = 12 set k = x + 1 7^k + 7^(k - 2) = 12 7^x + 7^k/7^2 = 12 7^k + 7^k/49 = 12 49·7^k/49 + 7^k/49 = 12 50·7^k/49 = 12 50·7^k = 12·49 7^k = (12·49)/50 7^k = 11.76 recall: k = x + 1 7^(x + 1) = 11.76 7·7^x = 11.76 7^x = 11.76/7 7^x = 1.68 x = ln(1.68)/ln(7) --- /// final result: ■ x = ln(1.68)/ln(7) --- /// check: 7^(ln(1.68)/ln(7) + 1) + 7^(ln(1.68)/ln(7) - 1) = 12 🙂
Nice Alternative
[7^(x+1)]-[7^(x-1)]=12 7×(7^x)+(7^x)/7=12 (48/7)(7^x)=12 (4/7)(7^x)=1 --> 4[7^(x-1)]=1 7^(x-1)=¼ Take logarithm: (x-1)log(7)=-log(4) --> x=1-⁷log(4) where ⁷log(4) is log based on 7
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Correction: Line 3 must be (50/7)(7^x)=12 7^(x-1)=6/25 Then take logarithm
LHS=3^(½+¼+⅛) =3^⅞ RHS=27^x =3^(3x) Therefore 3^⅞=3^(3x) --> 3x=⅞ x=7/24 Too easy for Math Olympiad.
3^(m+9)+3^(m+14)=244 [3^(m+9)](3⁵+1)=244 244[3^(m+9)]=244 --> 3^(m+9)=1 m+9=0 m=-9
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Sans passer par log 5
5^x=25=5^2 d0nc x=2
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x = (3log343(7)) / 2 = .5
2.
2× 5^x = 2 × 5^2 Therefore, x = 2
Thank you for the comments.
6x + 9x = 4x + 2 6x + 9x - 4x = 2 11x = 2 x = 2 / 11 x = 0.18181
Way too many steps for a simple equation like this.
It explains better
Exactly 💯
8^x + 2^x = 350 (2^3)^x + 2^x = 350 (2^x)^3 + 2^x = 350 let b = 2^x b^3 + b = 350 part proving 350 = 7^3+7 b^3 + b = 7*50 b^3 + b = 7*(49+1) b^3 + b = 7*(7^2+1) b^3 + b = 7*7^2+7 b^3 + b = 7^3+7 b = 7 2^x = 7 x = log_2(7) or log(7)/log(2)
Good alternative ❤
2^x=3 2 əsasdan loqarifmləsək loq 2^x = loq3 buradan xloq2= loq3 x=loq3 8^x+2^x=(2^3)^x+2^x=2^3x+2^x= =2^3loq3 +2^loq3=2^loq3^3+2^loq3= =2^loq27+2^loq3 =27+3=30 (loqarifmlər 2 əasdandır) 2-ci üsulla həllin doğruluğunun yoxlanması: 2^x=3 olduğundan 8^x+2^x=30 tənliyində 2^x-in yerinə 3 yazsaq (2^x)^3+2^x=3^3+3=27+3=30 dəqiq olardı. Təşəkkürlər.
Nice alternative
Very simple subscribe for more
Simple and nice steps to follow