Solving exponential equations by logs

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To solve the exponential equation 2^(2x) + 2^(x) - 12 = 0, it can be transformed by letting u = 2^x, leading to the quadratic equation u^2 + u - 12 = 0. This equation can be factored into (u - 3)(u + 4) = 0, yielding solutions u = 3 and u = -4. Since u represents 2^x and must be positive, only u = 3 is valid. Thus, solving for x gives x = log2(3). Understanding the properties of exponential and logarithmic functions is crucial for solving such equations.
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Homework Statement



2^(2x) + 2 ^(x) - 12 = 0

Homework Equations



none really

The Attempt at a Solution



so I think what you have to do is factor it
so it would be like

(2^x- )(2^x + )

then you set the factor equal to zero and solve for x but I'm not sure how to factor it.
 
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let u=2^x

now can you factor it?
 
Also notice that 2^2x = (2^x)^2
as previously mentioned u = 2^x, (u > 0, because the exponential function is always positive) so you get u^2 + u - 12 = 0, which is easy to solve. Just don't forget that u > 0.
 
also, it would serve you well to remember the graph of "e" and "ln"
 
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