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

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To solve the equation 2^(2x) - 2^(x) - 6 = 0, the substitution y = 2^x simplifies it to the quadratic equation y^2 - y - 6 = 0. This factors into (y - 3)(y + 2) = 0, yielding solutions y = 3 and y = -2. Since y = 2^x must be positive, only y = 3 is valid, leading to 2^x = 3. Finally, solving for x gives x = log2(3).
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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)
 
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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.
 
Karma said:
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! 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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