What is the logic behind this?? how do you solve this?

(2^(x) \- 2^(-x))/(2^(x) \+ 2^(-x)) = 1/3

3(2^(x) \- 2^(-x)) = (2^(x) \+ 2^(-x))

3\*2^(x) \- 3\*2^(-x) = 2^(x) \+ 2^(-x)

2\*2^(x) \- 4\*2^(-x) = 0

2^(x+1) \- (2^(2))\*2^(-x) = 0

2^(x+1)\- 2^(2-x) = 0

2^(x+1) = 2^(2-x)

(2^(x+1))/(2^(2-x)) = 1

2^(x+1-2+x) = 1

2^(2x-1) = 1

(2^(2x))/2 = 1

2^(2x) = 2

2x = 1

x = 1/2
Let’s make this a little bit easier by defining y=2^x. Then your problem reads

(y-1/y) / (y+1/y)=1/3

Normally, multiplying the top and bottom of a fraction by a variable is dangerous if the variable can be 0. But since 2^x can never be 0, we can do this with impunity. So, we multiply top and bottom by y to get.

(y^2 -1) / (y^2 +1)=1/3

Now it’s just a game of algebra. We multiply both sides by 3(y^2 +1) to remove the fractions and then solve:

3(y^2 -1)=(y^2 +1)

3y^2 -3=y^2 +1

2y^2 =4

y^2 =2

y=sqrt(2)=2^(1/2)

In that last line, I used the fact that taking the square root is the same as raising both sides to the 1/2 power. Now we use our definition for y to get

2^x =2^(1/2)

By inspection (or logarithms), x=1/2
I multiplied the top and bottom of the left side by 2^x so it simplified to (2^2x  -1)/(2^2x+ 1).  Then I cross multiplied and added 3 to both sides.
Now rewrite the 4 in the equation as 2^2 and you have 2^2x + 2^2 = 3(2^2x).
Divide both sides by 2^2x and you get:
1+2^(2-2x) = 3.
Subtract 1 from each side.
If 2^(2-2x) = 2^1  that means:
2-2x = 1. Subtract both sides by 2, divide by -2.
X = 1/2
Get the numerator and denominator on opposite sides of the equation, and work until you no longer have fractions.

That should be a good start
use componendo divideno

if a/b=c/d then (a+b)/(a-b)=(c+d)/(c-d)

set 2^x=a and 2^-x=b and you'll get the relation a=2b which you can solve easily
by
Use conjugates.
If a/b=c/d, then (b+a) /(b-a) =(d+c) /(d-c)
Whole equation changes to (2^x) /(2^-x) = 4/2
2^2x = 2
X=1/2
Just put 2^x as t and solve as a polynomial then you'll find t=2^0.5 thus x=0.5

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