Problem Set 1, 1H-3 (c), Can someone help me with this? The solution y=[e^x+-(e^x^2+4x^x)^1/2]/2 Is there a straight forward way to deduce that y-1<0? Thanks a lot!

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Problem Set 1, 1H-3 (c), Can someone help me with this? The solution y=[e^x+-(e^x^2+4x^x)^1/2]/2 Is there a straight forward way to deduce that y-1<0? Thanks a lot!

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You have mistakenly written < 0, it's y-1 > 0 Deduced like this \[2\ln y = \ln{(y+1)+x}\] \[\rightarrow 2\ln y - \ln(y+1)= x\] \[\rightarrow \ln y^2- \ln(y+1) = x\] \[\rightarrow \ln(\frac{y^2}{y+1})=x\] exponentiate \[\frac{y^2}{y+1}= e^x\] Now because e to the x is ALWAYS greater than ZERO we have \[e^x > 0\] so then must \[\frac{y^2}{y+1}> 0\rightarrow y^2>0\rightarrow y>0\] Now if \[y>0\] and \[e^x > 0\] then ( and this is important ) \[e^x \cdot y > e^x\] So divide both sides by \[e^x\] \[y > 1 \rightarrow y -1 > 0\] And thus we have shown that infact \[y-1 > 0\]
Thank you so much! It's the multiplying both side by e^x that I didn't know! Great help!
Sorry to bother you again: but how is it that e^xy>e^x again? if 0

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if \[e^x > 0\] and \[y>0\] and also from the equation we are given \[y^2 = e^xy + e^x\] we can deduce that y is greater than e^x because it takes something (positive) to be added to e^xy to EQUAL y squared. So if \[y>e^x\] it means that \[e^xy > e^x\] because ANYTHING multiplied by something greater than itself will be greater than itself.
Sorry about that, should have thought of it by myself, feeling really dumb right now!
Never let math make you feel 'dumb', it all depends on where your mind is at in a moment in time. This is why mathematics is so complicated, your expected to take so many many things into account all at once! However its moments like these that help some things 'stick' and for it to always be in the forefront of your memory. Never has it been more true than with mathematics, the saying 'we learn by our mistakes'! And remember that 'forgetting' is NOT 'dumb' or 'stupid', it's never being able to 'understand' something that is the problem!

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