I did it by a much quicker method which also answers your point about how many roots there are. 1. By inspection, x=2 is a solution. 2. 3^x is monotone increasing in x 3. x is monotone increasing in x 4. Therefore the left side is monotone increasing in x 5. Therefore x=2 is the only solution (assuming x is a real number, not a complex one).
x=2. Simples! Cambridge University would never have this as an entrance question unless they placed additional conditions as to how it was to be solved. I used the simplest method - by observation.
Honestly, x=2 and must be only 2 because above 2 it is too large and below 2 it is too small and it can't be negative or fractional or imaginary for the same reason, which I'm pretty sure is all you have shown, in an obfuscated fashion.
A nice tutorial in how to arrange two expressions into a form that can be reduced by the W function. Of course, you don't actually need to know about W, do long as you can rearrange both sides into the form (something) e^(something) and then equating the somethings. There was still an element of guess and check by knowing to break 11 into 9 and 2. Alternatively, guess and check quickly gives x=2 and it's siimple to show that the increasing function has just one solution.
I did it by a much quicker method which also answers your point about how many roots there are.
1. By inspection, x=2 is a solution.
2. 3^x is monotone increasing in x
3. x is monotone increasing in x
4. Therefore the left side is monotone increasing in x
5. Therefore x=2 is the only solution (assuming x is a real number, not a complex one).
x=2. Simples! Cambridge University would never have this as an entrance question unless they placed additional conditions as to how it was to be solved. I used the simplest method - by observation.
ur method is wrong cuz u cant probe that there is not any solution other than x=2
The question only asked me to solve it. It did not ask me to find every solution.
@asbarker31
most statements ask u to find possible values of x
@@outverse_edits3^x+x-11 is always growing when x>0, x=0 isn't a solution, and always negative when x
3^x is an increasing function as x increases. 11-x is a decreasing function as x increases. So there can be only one solution in the real numbers.
Great demo of method - thank you! Very helpful!
Honestly, x=2 and must be only 2 because above 2 it is too large and below 2 it is too small and it can't be negative or fractional or imaginary for the same reason, which I'm pretty sure is all you have shown, in an obfuscated fashion.
Is the point of this to find the most elaborate way possible to get a simple answer?
The solution of x=2 is obvious.
That is not the point of this video.
The point is about using the Lambert W function in appropriate situations.
A nice tutorial in how to arrange two expressions into a form that can be reduced by the W function.
Of course, you don't actually need to know about W, do long as you can rearrange both sides into the form (something) e^(something) and then equating the somethings. There was still an element of guess and check by knowing to break 11 into 9 and 2.
Alternatively, guess and check quickly gives x=2 and it's siimple to show that the increasing function has just one solution.
Great solution!!
.the best math channel in TH-cam
The solution is x = 2. Indeed, 3^2 = 9 and 9 + 2 = 11.
{3x+3x ➖ }+{x+x ➖ }={6x^2+x^2}=6x^4 3^3x^2^2 3^1x1^2 3x^2 (x ➖ 3x+2).
There is many theories to solve it.but one could very easily rea ch the solution by putting x=2.
How is that suppose to be hard
Can it be solved by another method?
Guess and check. Then demonstrating there can be only one solution.
👍👍
Impressive!!!
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Nul
Impressive!!!