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Logarithms Calculus help!

  • Thread starter Karma
  • Start date
  • #1
76
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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...

but the answer is log2(3)
 

Answers and Replies

  • #2
382
2
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.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Karma said:
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...

but the answer is log2(3)
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.
 
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