Logarithms Calculus help!

1. Sep 18, 2005

Karma

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

im realy lost in this class i just came for one day and the teacher said just try the question using logarithms :S and i dont wahts going on...

this is what i did

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

2. Sep 18, 2005

Leong

Let 2^x = y.
Solve the equation.
Use logarithms to solve for x.

3. Sep 19, 2005

HallsofIvy

Staff Emeritus
You "lost" the exponentials! 22x is equal to 4x but 4x is not 4x and 2x is not 2x!

As Leong suggested, since 22x is also (2x)2, let y= 2x 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= 2x= 3 and can solve for x using logarithms.
Of course 2x= -2 is impossible- a positive number to any power is never negative.

Last edited: Sep 19, 2005