A Nice Exponential Equation

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[yoav]

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<0 it is negative and it is 0 for x=0 the minimum point (0, 3).

[broytingaravsol]

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

[morteza]

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;

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;

Use the properties of the logarithm and simplify it;
(1)

I named this equation equation one
Multiply "a" on both sides of equation 1

I call this equation equation two;
So multiply equation 1 by negative 1 and add it to equation 2;

Get a factor from factor a-1

So a=1
As result x=ln(a)/ln(2)=ln(1)/ln(2)=0

[-basicmaths]

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

[martinfranek]

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. :)

[omsingh]

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

[pranavupadhyay]

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

[scottleung]

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

[muneeb]

You can also use graphical method to solve for x

[monishrules]

Which program do you use for the black board

[christopherellis]

It's totally bizarre that Americans say fourths for quarters

[xenumi]

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?

[ytlongbeach]

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.

[srividhyamoorthy]

Ehhh I was barely on time for this one

[tulio]

I think this video is repeated because I've seen it on your channel before