How do you solve: e^(2x)-e^(x)-6 = 0? I know the answer is x = ln(3) but I have no idea how to get there.
by 
Philip_P
ANSWERS: 2
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The trick is to realize that e^2x = (e^x)^2. So if you choose variable Q = e^x, then the equation reduces to Q^2 -Q -6 = 0 which is a quadratic you can solve (it factors nicely) for Q. Note that only positive roots are valid here, because e^x can't be zero or negative. Since Q = e^x, once you have the value of Q then x = ln(Q)
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first off we assume that e^x as x: e^(2x)-e^(x)-6 = 0 x^2-x-6=0 (x-3)(x+2)=0 so you have x=3 and x=-2 correct?. Now you substitute x with e^x as we did above. so you have e^x=3 => x=ln3 and e^x=-2 => x=ln-2 (which technically does NOT exist) Therefore you only take ln3 as the answer. And there you have it. Hope it helped :DD
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