He did put context for the function. The value of the W function wouldn't be written under normal circumstances, it is fine to just leave it as a function. But in that regard aren't solutions like logarithms and exponentials also difficult to do by hand? You use a calculator.
In this problem, we should draw two graphs: the graph of y=2^x and the graph of z=2x. We will see that the two graphs will intersect at two points and we deduce that the points are x=1andx=2. This is a quick solution to multiple choice questions.
There is no intersection because the graph "z=2x" doesn't pass through the y-axis unless we consider it a plane in which case it passes only through y=1, x=z=0.
Without watching: exponential = linear, that's a case for Lambert's W function. Just invert the sides, giving you 2^(-x) = 1/2 x^(-1). Multiply by x, giving you x 2^(-x) = 1/2. Negate the sides, giving you -x 2^(-x) = -1/2. Realise that 2 = e^ln2, giving you -x (e^ln2)^(-x) = -x e^(-x ln2) = -1/2. Multiply by ln2: -x ln2 e^(-x ln2) = -ln2 / 2. Now, it's the time for Lambert: W(-x ln2 e^(-x ln2)) = -x ln2 = W(-ln2 / 2). So x = -W(-ln 2 / 2)/ln2. These are easy, when you know W...
Loved you using the "Lambert Function" sledgehammer again but in a timed exam, I would have observed x=1 and x=2 as the solutions (since 2^x > 0 for all x) in a few seconds and moved to the next problem, but where's the practice and fun in that? Thanks for another morning coffee wake up problem & solution!
As an undergrad. mechanical eng. student, I didn't know about this function. But if I had this question on an exam, I'd argue that exponential = linear only have three possible solutions, right? 0, 1 or 2 solutions. If I know 2 x which are soluctions ofr this equation, there are no more x's which solves this equation.
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Deeply unsatisfying. Lots of tiny steps taking forever, then introduces the W function without context and pulls the solution of the air.
x₁ = -W(-ln(2)/2)/ln(2) = -W(-ln(2)∙(2^(-1))/ln(2) = -W(-ln(2)∙e^ln(2^(-1))/ln(2) = -W(-ln(2)∙e^(-ln(2))/ln(2) = = -(-ln(2))/ln(2) = 1
x₂ = -W(-2∙ln(2)/(2∙2))/ln(2) = -W(-2∙ln(2)/(2^2))/ln(2) = -W(-2∙ln(2)∙2^(-2))/ln(2) = -W(-2∙ln(2)∙e^ln(2^(-2)))/ln(2) =
= -W(-2∙ln(2)∙e^(-2∙ln(2)))/ln(2) = -(-2∙ln(2))/ln(2) = 2
Welcome to math
He did put context for the function. The value of the W function wouldn't be written under normal circumstances, it is fine to just leave it as a function. But in that regard aren't solutions like logarithms and exponentials also difficult to do by hand? You use a calculator.
Few calculators allow you to readlily use the W function, for the contrast.
In this problem, we should draw two graphs: the graph of y=2^x and the graph of z=2x. We will see that the two graphs will intersect at two points and we deduce that the points are x=1andx=2. This is a quick solution to multiple choice questions.
Yes, if you don't need answers with complex numbers
There is no intersection because the graph "z=2x" doesn't pass through the y-axis unless we consider it a plane in which case it passes only through y=1, x=z=0.
10 mn for.... THAT....??!!!!!!!! 😂😂😂😂😂😂 No INTEREST AT ALL, please dont waiste our time, and stop thinking you could bé a genius of maths !!!!!!
Without watching: exponential = linear, that's a case for Lambert's W function. Just invert the sides, giving you 2^(-x) = 1/2 x^(-1). Multiply by x, giving you x 2^(-x) = 1/2. Negate the sides, giving you -x 2^(-x) = -1/2. Realise that 2 = e^ln2, giving you -x (e^ln2)^(-x) = -x e^(-x ln2) = -1/2. Multiply by ln2: -x ln2 e^(-x ln2) = -ln2 / 2. Now, it's the time for Lambert: W(-x ln2 e^(-x ln2)) = -x ln2 = W(-ln2 / 2). So x = -W(-ln 2 / 2)/ln2.
These are easy, when you know W...
At first sight (without watching) there are 2 real solutions: x =1 and x = 2.
I guess there are additional complex solutions.
I think what you mean is ↄc=1 ↄc=2
Simple method -
2^x =2x
2^x = 2¹ * x¹
2^x = 2¹ or 2^x = x¹
Therefore,
x will be 1 or 2.
CHECK :
2^x = 2x
2¹ = 2*1
2=2
Similarly,
2^x = 2x
2^2 = 2*2
4=4
Therefore ,
(x=1,x=2)
Loved you using the "Lambert Function" sledgehammer again but in a timed exam, I would have observed x=1 and x=2 as the solutions (since 2^x > 0 for all x) in a few seconds and moved to the next problem, but where's the practice and fun in that? Thanks for another morning coffee wake up problem & solution!
2(2)=2^2 2^1=2(1) x=1 x=2
I think what you mean is 2(2)=2^2 2^1=2(1) ↄc=1 ↄc=2
Why is there a ten minute video about this
@@marcelhelayel1151 Because a lot of times the answer will be very complicated
x=1;2
Easy 2^X = 2X, Xlog2 = 2X, Xlog2 - 2X = 0, X(log2-2) = 0, X=0. Error Xlog2/2X = 1, log2/2 = 1 errror.
Watching your videos one might think the most important thing in mathematics is the Lambert W function.
It’s in my head.
2x (x ➖ 2x+2).
Can we not just use observation and substitution? This was solved in less than a minute using observation and substitution.
x2⁻ˣ = 1/2
-x2⁻ˣ = -1/2
-xln2e⁻ˣˡⁿ² = -(1/2)ln2
-xln2e⁻ˣˡⁿ² = (1/2)ln(1/2)
-xln2 = ln(1/2)
xln(1/2) = ln(1/2) => *x = 1*
but (1/2)ln(1)2) = (1/2)²(2)ln(1/2)
= (1/2)²ln[(1/2)²]
so -xln2e⁻ˣˡⁿ² = (1/2)²ln[(1/2)²]
-xln2 = ln[(1/2)²]
xln(1/2) = 2ln(1/2) => *x = 2*
Why on earth introduce a complex function and long explanation, when "by inspection" works well for this particular example?
As an undergrad. mechanical eng. student, I didn't know about this function. But if I had this question on an exam, I'd argue that exponential = linear only have three possible solutions, right? 0, 1 or 2 solutions. If I know 2 x which are soluctions ofr this equation, there are no more x's which solves this equation.
0 is not one of the solutions because 2^0=1 but 2*0=0
Awsome video 🎉🎉 love you ❤❤
Took 10 seconds to solve - 1 and 2. Why create 10 minute video for this?
To give an explaination, at exam you will have 0 points for not giving it, but yes, thats dumb
Because your approach omits other roots.
To show a simple example of how to solve this type of equation.
Try to calculate 3^x=5x in 10 seconds.
Too easy.
ↄc
For me that's one of Higher Mathematics' signature moves that make this channel's video stand out from the others
X=2
I think what you mean is ↄc=2
solve for ↄc
hi may i know how to contact you？
how is nobody talking about the ↄc