(1) 2^p + 2^q + 2^r = 42 We can consider the sum (1) as a binary number in the notation of which there are 3 ones. power 2: 7654 3210 42 = 0010 1010 42 has exactly three ones in its binary notation Therefore p = 5, q = 3, r= 1
Solving equations and their systems in integers generally leads to limiting the number of combinations to check. In this problem, we are unlucky that 30 has so many divisors, but we have a different favorable circumstance. The equation represents a hyperbola passing through the point (0, 0). Given the form of the problem, we reject the solution (0,0). It also has asymptotes: a horizontal asymptote at y = 5/7 and a vertical asymptote at x = 6/7. From this (it's enough to sketch the hyperbola), it follows that the solutions are pairs where x and y are positive, and, as can be easily checked, one of them must be equal to 1. It is sufficient to check by substituting x = 1, which gives y = 5, and x = 2. For x = 2, the value of y is less than 2, and the function is positive and decreasing, so the only option for an integer solution is y = 1. This gives x = 5.
(1) 5/a + 6/b = 7 (a,b from Z+) Obviously 5/a + 6/b < 7 when a > 5 (a,b from Z+) Therefore a can be (2) a = 1,2,3, 4 or 5 From (1) 6/b = 7 - 5/a = (7a - 5) / a => b = 6[a/(7a - 5)] Test all a from (2) a 6[a/(7a - 5)] b Result 1 6[1/(7*1 - 5] = 6/2 = 3 3 (1,3) solution 2 6[2/(7*2 - 5] = 12/9 - no solution 3 6[3/(7*3 - 5] = 18/16 - no solution 4 6[4/(7*4 - 5] = 24/23 - no solution 5 6[5/(7*5 - 5] = 30/30 = 1 1 (5,1) solution
@2012tulio3 сағат бұрын
X={-0.6826}
@2012tulio3 сағат бұрын
{p,q,r}={1;3;5}
@shrikrishnagokhale35574 сағат бұрын
Sorry,instant answer is wrong.Will try seriously.
@shrikrishnagokhale35574 сағат бұрын
Instant answer is x=1
@user-nz1fc6ml3s4 сағат бұрын
42=(binary)101010=2^5+2^3+2^1 hence r=5, q=3, p=1
@vultusalbus42164 сағат бұрын
Thank you very much for explaining how to solve difficult equations. I can find the solution to an equation more easily than I can write down the proof with steps, like some gifted people who just guess the answers but can’t explain it with proof
@superacademy2474 сағат бұрын
You're welcome 🙂. Glad you liked it. 🙏🙏🙏
@murtadah63275 сағат бұрын
I think not much time is provided for solving such simple problems. Hence, hit and trial will be the best technique to adopt, as suggested by others here.
@superacademy2475 сағат бұрын
Hit and trial could be a guide into the solution but if you want to demonstrate authority over the subject you ought to show your workings vividly.
Too long solution, we can see a the blink of eye that a=5,b=1 and/or a=1,b=3
@boguslawszostak17846 сағат бұрын
A blink of an eye is enough to find one or two solution. But with just a blink of an eye, you won't prove that it is the only solution.
@TommyRaines5 сағат бұрын
Ridiculous rigmarole when you can solve this just by looking at it !
@TommyRaines5 сағат бұрын
@boguslawszostak1784 well, the blink of my eye produced all the solutions. Imagination and insight are valuable skills, anyone (and computers) can grind out laborious "brute force" methods but what do they really learn ?
@vultusalbus42164 сағат бұрын
@@boguslawszostak1784Gifted people often guess the answer but can hardly explain it with proof. If I say that log(log)10 billion equals 1, it’s not enough if I can’t explain it. Same for log(log)(base2) of 65536. How do I prove it’s 4 without writing down all the steps ? Because that’s how one avoids getting an F in math class
@emmanuellaroy28684 сағат бұрын
I saw an other visible solution : a=1 / b= 3
@anniesworldeverythingsurro82146 сағат бұрын
X = - 0.2 only
@SidneiMV9 сағат бұрын
4:40 wrong value for x correct value is *W(-5ln5)/(-5ln5)*
@superacademy2478 сағат бұрын
It's exactly 💯 what I have done ✅
@SidneiMV9 сағат бұрын
x = ± 5⁵ˣ [ I ] x = 5⁵ˣ x5⁻⁵ˣ = 1 -5x5⁻⁵ˣ = -5 (-5xln5)e⁻⁵ˣˡⁿ⁵= -5ln5 -5xln5 = W(-5ln5) *x = W(-5ln5)/(-5ln5)* [ II ] x = -5⁵ˣ -x = 5⁵ˣ -x5⁻⁵ˣ = 1 -5x5⁻⁵ˣ = 5 (-5xln5)e⁻⁵ˣˡⁿ⁵= 5ln5 -5xln5 = W(5ln5) = ln5 *x = -1/5*
@superacademy2478 сағат бұрын
Awesome 😎💯
@user-vi8dh7gv3t10 сағат бұрын
125.
@soojin112 сағат бұрын
72-21|7/4 is not 21(7)-82/4 is. 😅😂 all not, but 203(7)-533/16 is!! You are always truth!!
@superacademy24711 сағат бұрын
Thanks 😊 for correction. I saw it but I knew it was an error/typo. Thanks again for coming out to clarify 👌💕
@ronaldnoll324717 сағат бұрын
This is not a simplification but a complication
@superacademy24717 сағат бұрын
The denominator is rational. Which is less complicated
@peterotto71219 сағат бұрын
A bullshit Test delivers bullshit results.
@lasalleman679219 сағат бұрын
I don't see how x^7= [203(7)^.5]/16 - 533/16 represents a simplification of the original statement.
@prollysine19 сағат бұрын
we get , x^4-40x^2-x+380=0 , add 5x^3 , -5x^3 , by fakt. , (x-5)(x^3+5x^2-15x-76)=0 , solu. , x=5 , test , sqrt(20+sqrt(20+5))=sqrt(20+5) , --> x>0 , --> V25=5 , x=5 , same , OK , /// is not part of the basic equation x^3+5x^2-15x-76=0 , (x+4)(x^2+x-19)=0 , x= -4 , x^2+x-19=0 , x=(-1 +/- V77)/2 , x= -4 , (-1+V77)/2 , (-1-V77)/2 , not a solution to the basic task /// ,
@SanePerson120 сағат бұрын
You're working too hard, in my opinion. Let x = (√7 - 1)/2. Compute x², which can be written as x² = (3/2) - x. Now just recursively use that formula: x³ = x•x² = x•[(3/2) - x] = (3/2)x - x² = (3/2)x - ((3/2) - x) = (5/2)x - 3/2. Now just compute x⁷ = x(x³)² = x•[(5/2)x - 3/2]² then substitute back in for x after doing the straightforward simplification.
x+y=1; i,e x=1-y ; 3^(1-y)-3^y=3 ; i,e [3/(3^y)]-3^y-3=0 ; Let a=3^y ; i,e 3-a^2-3a=0 ; a^2+3a-3=0 ; i,e a= (-3+√21)/2 or a= (-3-√21)/2 ; i,e 3^y= (-3+√21)/2 ; i,e y= [log (√21-3)-log2]/log3 or y= [log(-√21-3)-log2]/log3 & x=[log3-log(√21-3)+log2]/log3 or x= [log3-log(-√21-3)+log2]/log3. x and y each have an imaginary value.
@igorgoliney949421 сағат бұрын
Looking at the thumbnail for about half a minute, one concludes that x= 5 is the solution. The uniqueness is more difficult. You see, x cannot be much larger than sqrt(20) >4. let's try 5, es stimmt.
@user-vi8dh7gv3t22 сағат бұрын
Ответ 24.
@superacademy24722 сағат бұрын
Take your time
@ronaldnoll324723 сағат бұрын
I once made a graphical solution using Excel and came up with the following solutions: X(1) = -0.435 and x(2) = 0.768
@RyanLewis-Johnson-wq6xsКүн бұрын
5^(10x)=x^2 x=-1/5=-0.2 x = -(W_n(-5 log(5)))/(5 log(5)), n element Z
@tgeofreyКүн бұрын
We can use complex number formula
@2012tulioКүн бұрын
Square both sides and arrange you will get Sqrt(20+x) = x^2-20 Let a = sqrt (20+x) So a^2 = 20+x Let b= x^2-20 a-b=0-----eq1 a^2+b=x^2+x ----eq2 eq1+eq2: a^2+a=x^2+x By comparison: a^2=x^2 and a=x So sqrt(20+x)=x 20+x =x^2 x^2-x-20=0 By solving this quadratic equation you will get x= 5 or x=-4 but after verifying , x=5 is the valid solution.
@superacademy247Күн бұрын
This is a powerful alternative method. Thanks 👍💯 for your input and support. 🙏🙏🙏