The problem I'm having with this explanation is that it doesn't make intuitive sense to let 1/x = -t. I tried it by substituting 1/x= t and it turned into a mess. Why would someone think to let 1/x be equal to NEGATIVE t?? It works, but what's the intuition there, other than the fact that you knew it beforehand.
1) the most important thing is to determine the substitution, which is 1/x=t. 2) having in mind that we have (-x), in order to get rid of this ‘-‘ and to ‘convert’ it into something known and useful [i.g. a^a or (1/a)-1)^a], the substitution has to be negative, i.g. 1/x=-t
Nice solution, but you have a couples typo or copy errors 3. line should be 1 = -x/4^x = -x*4^-x (not / but *) 4. line should be ln4 = -xln4*e^-xln4 (ln4 is missing on the right side) 5. line W(-x*e^-xln4) in not type W(ae^a) 1 = -x*4^-x ln4 = -xln4*4^-x 2ln2 = -xln4*e^-xln4 W[ln2*e^ln2] = W[-xln4*e^-xln4] ln2 = -xln4 x = -ln2/ln4 = -ln2/2ln2 = -1/2
Honestly, it's not a difficult question at all, in fact, I would say that any college student can solve this question. It's easy to see that 4^x, for x = - 1/2 results in (2²)^(-½) = 1/2. I found it very easy.
The problem I'm having with this explanation is that it doesn't make intuitive sense to let 1/x = -t. I tried it by substituting 1/x= t and it turned into a mess. Why would someone think to let 1/x be equal to NEGATIVE t?? It works, but what's the intuition there, other than the fact that you knew it beforehand.
its because he said so 💀🤣
The graph gives you an idea. You are right, it is not intutive.
1) the most important thing is to determine the substitution, which is 1/x=t. 2) having in mind that we have (-x), in order to get rid of this ‘-‘ and to ‘convert’ it into something known and useful [i.g. a^a or (1/a)-1)^a], the substitution has to be negative, i.g. 1/x=-t
Looking at the original problem, if x is real, then it must be 0.
Or sketching 4^x on the same axes as x, by observation, x
4ˣ = -x
u = -x => x = -u
4⁻ᵘ = u
-uln4 = lnu
(1/u)ln(1/u) = 2ln2
1/u = 2 => u = 1/2 => *x = -1/2*
4^x=-x taking ln , ln4^x=ln(-x) so
xln2^2=ln(-x) , 2xln(1/2)^-1 ,-2xln(1/2)=1ln(-x)
-2x=1 and1/2=-x result x=-1/2
Sir,
4^x+x=0
4^x=-x,
1=-x/4^-x,
Ln4=-x.e^-xln4,
W(Ln4)=w(-x.e^-xln4)
X=-W(Ln4)/Ln4
Ln4= 2Ln2,
X=-W(e^ln2.Ln2)/2ln2
-W(e^ln2.Ln2)=-ln2,
X=-ln2/2ln2
X=-1/2/=
Hi, how did you make line 4 from line 3, could you please help explain? thanks
@@yubichun4928 you should learn Lambart"W" function,
W(a.e^a)=a,
Nice solution, but you have a couples typo or copy errors
3. line should be 1 = -x/4^x = -x*4^-x (not / but *)
4. line should be ln4 = -xln4*e^-xln4 (ln4 is missing on the right side)
5. line W(-x*e^-xln4) in not type W(ae^a)
1 = -x*4^-x
ln4 = -xln4*4^-x
2ln2 = -xln4*e^-xln4
W[ln2*e^ln2] = W[-xln4*e^-xln4]
ln2 = -xln4
x = -ln2/ln4 = -ln2/2ln2 = -1/2
@@gregorymagery8637 Thanks you sir, describe my errors ....❤️
Honestly, it's not a difficult question at all, in fact, I would say that any college student can solve this question. It's easy to see that 4^x, for x = - 1/2 results in (2²)^(-½) = 1/2. I found it very easy.
C'mon, this is -1/2 by inspection.
Using Lambert W
4^x + x = 0
4^x = -x
x*ln4 = ln(-x) |*-1
(-x)*2ln2 = -ln(-x) = ln(-x)^-1 = ln(1/-x)
2ln2 = (1/-x)*ln(1/-x)
W[2ln2] = W[(1/-x)*ln(1/-x)]
W[e^ln2*ln2] = W[e^ln(1/-x)*ln(1/-x)]
ln2 = ln(1/-x)
=> 2 = 1/-x
x = -1/2
{4x+4x ➖ }+{x+x ➖ }={8x^2+x^2}=8x^4 2^3x^2 1^1^1x1^2 1x^2 (x ➖ 2x+1).
No
Puh, I just guessed right
everyone can guess.. but everyone can't solve....
x = -1/2