in another way log 4^x=log x^64 x log 4 = 64 log x (1/64)log 4= (1/x) log x log 4 ^(1/64) = log x^(1/x) 4^(1/64) = x ^(1/x) 4^((4/4).(1/64)) = x ^ (1/x) 4^(4/256)=x ^(1/x) 256^(1/256) = x ^ (1/x) by analogy L /R X = 256
4=100 in binary & 64= 1000000 just adding 1000000"00"=256 !!! impressive no ??? so if (4)^6=4096 8^(4)=4096 trying something 6=110 and 8=1000 thinking 100 (sub cause not 00 but 10 & not testing with 11 or 01 !) 100=4 yes i found it again ! (with non-conventionnal practise !) XD
Drawing the graphs of f(x) = x^(1/64) and g(x) = x reveals that there is another solution, with x approximately equal to 1.0. Using iteration the approximation can be refined to 1.022393. I don't know how this solution could be derived algebraically.
You can use the Lambert W function to find the second real solution. In fact, there are three solutions. x = -0.979017 (approximately) is a third solution.
I guess yoy mean f(x)=4^(x/64). Original you have 4^x=x^64>0. You can write 4^(x/32)=x^2 (also both >0) without a problem. But if you want to go 1 step farther then you have to use |x| , so the correct is 4^(x/64)=|x| and the functions are f(x)=4^(x/64) , g(x)=|x|. Drawing these graphs you can see all 3 real solutions.
@@Misha-g3b 4^x=x^64, we power both sides by 4=>256^x=x^256=>x/ln(x)=256/ln(256)=46.1662413084=>we suppose y=46.1662413084 then ln(x)=e^(x/y)=> -1/y=(-x/y)e^(-x/y) , now we can use Lambert W Function=> W(-1/y)=-x/y => x=e^(0.0221458994978)=1.02239294021 , in this case we have just real solution, but if y
yes ... that is why if x^x = y^y does not mean that x=y and also x^(1/x)= y ^ ( 1/y ) does not mean that x=y ... I see this often not concidered ...
x*=256. Other 2 roots are on intervals (-1, 0), (1, 2): x**=-0,97...; x***=1,02... .
in another way
log 4^x=log x^64
x log 4 = 64 log x
(1/64)log 4= (1/x) log x
log 4 ^(1/64) = log x^(1/x)
4^(1/64) = x ^(1/x)
4^((4/4).(1/64)) = x ^ (1/x)
4^(4/256)=x ^(1/x)
256^(1/256) = x ^ (1/x)
by analogy L /R
X = 256
Excelente!!!
Thankyou so much Boss.
Such lovey feedback means a lot.
Thanks 😊
example ❤
Thanks ❤️
both sides power to 4 then 256^x=x^256=>x=256
How...?
Using Which property of Exponents..?
@@MathBeast.channel-l9i 4^4=256, 64*4=256, (4^x)^4=(4^4)^x=256^x, (x^64)^4=x^(64*4)=x^256
4=100 in binary & 64= 1000000 just adding 1000000"00"=256 !!! impressive no ???
so if (4)^6=4096 8^(4)=4096 trying something 6=110 and 8=1000
thinking 100 (sub cause not 00 but 10 & not testing with 11 or 01 !)
100=4 yes i found it again ! (with non-conventionnal practise !) XD
Math Olympiad Question: 4ˣ = x⁶⁴; x =?
4ˣ = x⁶⁴ > 0, x > 0; (x⁶⁴)¹⸍⁽⁶⁴ˣ⁾ = (4ˣ)¹⸍⁽⁶⁴ˣ⁾, x¹⸍ˣ = 4¹⸍⁶⁴ = 4⁴⸍²⁵⁶ = (4⁴)¹⸍²⁵⁶ = 256¹⸍²⁵⁶
x¹⸍ˣ = 256¹⸍²⁵⁶; x = 256
Answer check:
x = 256: 4ˣ = 4²⁵⁶ = 4⁴⁽⁶⁴⁾ = (4⁴)⁶⁴ = 256⁶⁴ = x⁶⁴; Confirmed
Final answer:
x = 256
Nice Approach 👍
Actually, x can be negative in this case.
In fact, x = -0.979017 (approximately) is another solution.
@@davidbrisbane7206
Glad to know. Thanks,
Solve carefully 👀
2x-3=5
X!=?
Maybe, x!=?
Too easy
Not too easy.
@@Misha-g3b maybe not for you.
@@iceman9678 Then try this, if You want: x^x^5=100 => x=... (?).
Then try this, if You want: x^x^5=100 => x=... (?).
Prove that it is the only root
Drawing the graphs of f(x) = x^(1/64) and g(x) = x reveals that there is another solution, with x approximately equal to 1.0. Using iteration the approximation can be refined to 1.022393. I don't know how this solution could be derived algebraically.
Verify your solution by plugging into the original problem. Is 1.02^64 close to being equal to 4^1.02?
You can use the Lambert W function to find the second real solution. In fact, there are three solutions.
x = -0.979017 (approximately) is a third solution.
I guess yoy mean f(x)=4^(x/64). Original you have 4^x=x^64>0. You can write 4^(x/32)=x^2 (also both >0) without a problem. But if you want to go 1 step farther then you have to use |x| , so the correct is 4^(x/64)=|x| and the functions are f(x)=4^(x/64) , g(x)=|x|. Drawing these graphs you can see all 3 real solutions.
indeed there is at least 4 answers for this question, but some of them är complex
Write down a complex root.
@@Misha-g3b 4^x=x^64, we power both sides by 4=>256^x=x^256=>x/ln(x)=256/ln(256)=46.1662413084=>we suppose y=46.1662413084 then ln(x)=e^(x/y)=> -1/y=(-x/y)e^(-x/y) , now we can use Lambert W Function=> W(-1/y)=-x/y => x=e^(0.0221458994978)=1.02239294021 , in this case we have just real solution, but if y
Now, I have to save my mobile data for eng vs aus.
Which team were you supporting 🧐
all the same answer 256 raise 256??
The music Is horrible 😢😢
Sorry for that Boss 😔
Müziksiz olmasını tercih ederiz.
Okay Boss
Utter nonsense!!! There is another solution too
Yes there is too
Still 2 roots.
3 real solutions
Use the Lambert W function W(■*e^■) = ■
4^x = x^64
2^(2*x) = x^64
ln(2^(2*x)) = ln(x^64)
2*x*ln(2) = 64*ln|x|
x*ln(2) = 32*ln|x| ===> two cases
1st case: x > 0
x*ln(2) = 32*ln(x)
ln(x)*x^(-1) = ln(2)/32
ln(x)*e^ln(x^(-1)) = ln(2)/32
ln(x)*e^(-ln(x)) = ln(2)/32
-ln(x)*e^(-ln(x)) = -ln(2)/32
W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32)
-ln(x) = W(-ln(2)/32)
ln(x) = -W(-ln(2)/32)
x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977...
x₂ = e^(-W₋₁(-ln(2)/32)) = 256
2nd case: x < 0
x*ln(2) = 32*ln(-x)
ln(-x)*x^(-1) = ln(2)/32
-ln(-x)*x^(-1) = -ln(2)/32
ln(-x)*(-x)^(-1) = -ln(2)/32
ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32
ln(-x)*e^(-ln(-x)) = -ln(2)/32
-ln(-x)*e^(-ln(-x)) = ln(2)/32
W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32)
-ln(-x) = W(ln(2)/32)
ln(-x) = -W(ln(2)/32)
-x = e^(-W(ln(2)/32))
x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...
@@payoo_2674 Nice Approach 👍
😂😅
r/beatmetoit
The good mathematician is immediately visible.
Hi bro, math😉. Just wanted to write - there are 3 solutions, but You have ahead me