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[damjan]

So you basically didn't solve the equation but guessed the solution and were lucky. 7:12 Why did you jump to conclusion that x>=2 is not possible?

[DrQuatsch]

I am going to assume this is a Diophantine equation, otherwise it would be impossible to solve. If we inspect the equation, we can immediately see that x = 0 will be a problem. This gives us a 0/0 situation. So x = 0 cannot be a solution. Now we can rewrite the numerator as 8^x - 2^x = (2^3)^x - 2^x = (2^x)^3 - 2^x = 2^x * ((2^x)^2 - 1) = 2^x * (2^x - 1) * (2^x + 1). The denominator can be written as 6^x - 3^x = (2 * 3)^x - 3^x = 2^x * 3^x - 3^x = 3^x * (2^x - 1). Now we can rewrite the whole fraction as: [2^x * (2^x - 1) * (2^x + 1)]/[3^x * (2^x - 1)] = [2^x * (2^x + 1)]/3^x = (2/3)^x * (2^x + 1).
So we have the equation: (2/3)^x * (2^x + 1) = 2. Let's multiply LHS and RHS with (3/2)^x; 2^x + 1 = 2 * (3/2)^x = 3^x/2^(x - 1). As this is a Diophantine equation, the LHS is an integer for x > 0. This means the RHS also needs to be an integer. For x > 0 this only works for x = 1. If you take x > 1, we will have a fraction, because we are dividing by a number bigger than 1. If x < 0 we can substitute x = -y with y > 0. The LHS will be 2^(-y) + 1 = 1/2^y + 1 = (1 + 2^y)/2^y. The RHS will be 3^(-y)/2^(-y-1) = 2^(y+1)/3^y. This leads to (2^y + 1)/2^y = 2^(y+1)/3^y. Multiply both sides by 2^y to get: 2^y + 1 = 2^(2y + 1)/3^y. The LHS is an integer, so the RHS has to be one as well. 2^(2y+1) is a power of 2, and 3^y is a power of 3, so these will never be a multiple of each other if the powers are both bigger than 0, which in this case is true. So there are no solution for y > 0, which means there are no solutions for x < 0.
This leaves the only possible solution from before of x = 1. So let's check that in the original equation; [8^1 - 2^1]/[6^1 - 3^1] = (8 - 2)/(6 - 3) = 6/3 = 2. This checks out, so x = 1 is our solution.
If the equation is not a Diophantine equation, there's a multitude of possible answers in the complex plane alone.

[jcxmej]

If you were going to use substitution you could have done in the first equation itself. There was no need for solving it.

[ИванБелинский-кю]

on the left side with x >=2 we have the multiply of even and odd numbers. The result will be the even number. On left side 3 power x for x >= 1 will be always even number. So there are no solution for integer x > 1

[ИванБелинский-кю]

I cannot find out one thing - where is the requirements of X to be only integer for this equation? If there is such requirement - then the solution is correct. Otherwise I cannot find the solution that proofs that there is only one solution for this equation.

[manojchaugule]

very nice explaination👍
you are a angel mam for me

[kgs]

With all due respect ma'am, without pen and paper i got the answer as x=1..
I just saw the whole video for the right way to solve the problem, but no use.

[michellauzon]

Assuming x /= 0, the equation becomes (2/3)^x (2^x+1) = 2 . The best i can do is the following: x = 0 and x = 1 are two obvious solutions and they are the only one because the left side is strictly convex (long calculation). So x = 1 is the only solution.

[Michael_Alaska]

Chipped fingernail polish is distracting, it would just take a minute to use some nail polish remover.

[rajesh-dhdl]

Excellent way to solve .. however I feel logic x cannot be greater then 2 … should be substantiate with logic ..

Refer equation 2 power x( 2 power x +1) / 3 power x …

can be written as (4/3) power x plus (2/3) power x.

Sum of both term is 2 means both terms are less than 2 and as raise to x, so cannot be negative.

Since 4/3 is bigger than 1, so if x=3 then (4/3) power 3 = 64/27 >2. Thus x value cannot be 3 or more .

Now let us check x for 2.