Solving for X using Logarithms in Calculus: 2^(2x)-2^(x)-6=0 Explained

[Karma]

2^(2x)-2^(x)-6=0
solve for X..

im really lost in this class i just came for one day and the teacher said just try the question using logarithms :S and i don't wahts going on...

this is what i did

4x-2x-6=0
2x-6=0
x=3...

but the answer is log2(3)

 

[Leong]

quadratic equation and logarithms:
Let 2^x = y.
Form a quadratic equation.
Solve the equation.
Substitute your answer back into 2^x = y.
Use logarithms to solve for x.

 

[HallsofIvy]

2^(2x)-2^(x)-6=0
solve for X..

im really lost in this class i just came for one day and the teacher said just try the question using logarithms :S and i don't wahts going on...

this is what i did

4x-2x-6=0
2x-6=0
x=3...

but the answer is log2(3)

You "lost" the exponentials! 2^2x is equal to 4^x but 4^x is not 4x and 2^x is not 2x!

As Leong suggested, since 2^2x is also (2^x)², let y= 2^x so that your equation becomes the quadratic equation y²-y- 6= (y-3)(y+2)= 0 which has solutions y= 3, y= -2.
Now you know that y= 2^x= 3 and can solve for x using logarithms.
Of course 2^x= -2 is impossible- a positive number to any power is never negative.