Bulgarian Math Olympiad | Interesting Algebra Problem | Exponential Equation

In this video, I am presenting step-by-step solution to this interesting algebra problem and also the exponential equation from Bulgarian Math Olympiad. This question is in the rational function that we will solve for exponential equation, and pretty interesting algebra problem from Bulgarian Math Olympiad. I specifically used graphing of exponential function, and check the final solutions using them. Come check this video out and watch it until the end. Exponential equation is an important topic in algebra, and this algebra problem is from Bulgarian Math Olympiad you can enjoy watching. More to come! Stay tuned!

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[justinnitoi]

Start by factoring the numerator with a common factor of 2^x and the denominator with a factor of 3^x:
2^x(4^x - 1)/3^x(2^x - 1) = 2, x=/=0
2^x(2^x + 1)/3^x = 2
2^x(2^x + 1) = 2*3^x
2^(x-1) * (2^x + 1) = 3^x

Notice that when x>1, the left side will always be even and the right side will be odd.

When x = 1, it satisfies the equation.

When x < 0:

The equation can be rewritten as:
2^(-x-1) * (2^(-x) + 1) = 3^(-x), where x > 0.
(1/2)^(x+1) * ((1/2)^x + 1) = (1/3)^x
(1/2)^(2x+1) + (1/2)^(x+1) = (1/3)^x

Here, notice that the right side will always have an odd denominator and the left side will always have an even denominator.

Therefore, the only possible solution is: x=1.

[Min-cvnt]

This question is so lovely professor. like this so much

[GK_Susebron]

If x\in Z, aren't we supposed to consider x<0 as well?

[saqeefsanan]

2³*-2*/2*3*-3*=2
2*(2²*-1)/3*(2*-1)=2
(⅔)* ×(2*+1)=2
at x=0 and x=1 it satisfies the equation.

[rubensramos]

After some algebra one gets (4/3)^x+(2/3)^x = 2, whose solution is x = 1. The general analytical solution of the problem a^x+b^x = c^x can be found in the (very small) paper "On the Solutions of a^x + b^x = c^x", than can be downloaded on researchgate.

[yaboi]

I'm not a student and I don't like math, why did YouTube reccomend this? And why did I watch the whole thing lol

[joshmceowen]

I just looked at it and realized it works without the exponent.