A Nice Exponential Equation

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Күн бұрын

Пікірлер: 20

@yoav613 Жыл бұрын

Nice! For the second method with the derivative you can also write f' as ((3/4)^x-(2/3)^x)ln(3/2)+(2^x-(3/4)^x)ln2,and since 3/4>2/3 and 2>3/4 and ln(3/2) and ln2 are positive,for x>0 the derivative is positive and for x

@broytingaravsol Жыл бұрын

more with post-approaches if setting a=2/3, b=3/4, then a linear combination that ab=1/2, or 1/ab=2, as the base in 2^x

@morteza3268 Жыл бұрын

Well, we have to do it; 2^x+(2/3)^x+(3/4)^x=3=a+b+c What I did is that I divided the number 3 into 3 parts a, b and c, each of these 3 parts is equal to the parts from the left side of the equality as follows; 2^x=a so ×=ln(a)/ln(2) (2/3)^x=b so x=ln(b)/ln(2/3) (3/4)^x=c So x=ln(c)/ln(3/4) Make the x parts equal; ln(a)/ln(2)=ln(b)/ln(2/3)=ln(c)/ln(3/4) Use these equations as follows; Ln(a)/ln(2)=ln(b)/ln(2/3) Ln(a)/ln(2)=ln(c)/ln(3/4) Set both sides of each equation to base e; a^(1/ln(2))=b^(1/ln(2/3)) a^(1/ln(2))=c^(1/ln(3/4)) So write b and c in terms of a (leave b and c alone) b=a^(ln(2/3)/ln(2)) c=a^(ln(3/4)/ln(2)) And we know; a+b+c=3 So put b and c; a+a^(ln(2/3)/ln(2))+a^(ln(3/4)/ln(2))=3 Use the properties of the logarithm and simplify it; (1) a+a^(1-ln(3)/ln(2))+a^(ln(3)/ln(2)-2)=3 I named this equation equation one Multiply "a" on both sides of equation 1 a²+a^(2-ln(3)/ln(2))+a^(ln3)/ln(2)-1)=3a I call this equation equation two; So multiply equation 1 by negative 1 and add it to equation 2; a²-a+a^(-ln(3)/ln(2))(a²-a)+a^(ln(3)/ln(2)-2)(a-1)-3(a-1)=0 Get a factor from factor a-1 (a-1)(a+a^(-ln(3)/ln(2)+1)+a^(ln(3)/ln(2)-2)-3)=0 So a=1 As result x=ln(a)/ln(2)=ln(1)/ln(2)=0

@pranavupadhyay8181 Жыл бұрын

Somebody please tell me that where to use this AM-GM inequality...?

@-basicmaths862 Жыл бұрын

Nice example ! Substitute x=3a, 2^a=p,(2/3)^a=q,(3/4)^a=r Then p q r= 1 &p+q+r is not equal to Zero. p^3+q^3+r^3=3pqr ,p=q=r Means 2^a=(2/3)^a=(3/4)^a. This is possible for a=0. x=3a, x=0

@omsingh7683 Жыл бұрын

Hey sir I hope you are doing great in your life as a student I'm commenting down here because I have request to make that is please do some more videos on recursive sequence problems

@scottleung9587 Жыл бұрын

I got x=0 by guessing and checking. I tried doing deeper analysis after that, but to no avail.

@muneeb7875 Жыл бұрын

You can also use graphical method to solve for x

@martinfranek4568 Жыл бұрын

A very cool idea with the AG inequality. I liked it a lot, however I found a different way to solve the equation: 2^x = (2/3)^x = (3/4)^x Just take a look at 2^x = (2/3)^x Rewrite as 2^x - (2^x)/(3^x) = 0 Dividing by 2^x (which never equals 0) and multiplying by 3^x (the same) we get 3^x = 1 Which has only one solution: x = 0 because f(x) = 3^x is always increasing. Double checking, x=0 is indeed a solution to the original equation. Without the need of calculus. :)

@MrTeniguafez Жыл бұрын

Would substitution not work here?

@monishrules6580 Жыл бұрын

Which program do you use for the black board

@SyberMath Жыл бұрын

Notability

@xenumi Жыл бұрын

How do you suddenly declare x to be an integer 83% of the way through the solution? It was not stated or proved at the beginning, and I could not find it afterwards. Am I missing something?

@deltalima6703 Жыл бұрын

I found that fishy. Thumbs down just for that. What if x is complex? He never even looked at that, very disappointing.

@christopherellis2663 Жыл бұрын

It's totally bizarre that Americans say fourths for quarters

@ytlongbeach Жыл бұрын

i just look at this problem this way: * all three terms on the left are increasing functions, and the term on the right is a constant. hence, there can be 0 or 1 solutions max. * x=0 is an easy first guess, and there's one max solution, so we're done. is guessing x=0 bad math? i can't help it. it jumps right out at me.