solve the equation Watch

fadya
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#1
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#1
2(e^3x+1)=10
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swervyshawn
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#2
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X = ((ln5) -1) /3


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Phichi
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#3
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How about you solve it? Where's your thoughts or working?
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Phichi
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#4
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(Original post by swervyshawn)
X = (ln4)/3


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http://www.thestudentroom.co.uk/showthread.php?t=403989

'To make it perfectly clear, the consensus (and forum rule), is that posting of full solutions should be considered a last resort. If your first contribution to a thread is to post a full solution, the moderators are quite entitled to take action. '
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samsama
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(Original post by Phichi)
http://www.thestudentroom.co.uk/showthread.php?t=403989

'To make it perfectly clear, the consensus (and forum rule), is that posting of full solutions should be considered a last resort. If your first contribution to a thread is to post a full solution, the moderators are quite entitled to take action. '
He posted the answer, not the full solution. It's incorrect anyway.
2(e^3x+1)=10 is not the same as 2(e^(3x+1))=10
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Phichi
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(Original post by samsama)
He posted the answer, not the full solution. It's incorrect anyway.
2(e^3x+1)=10 is not the same as 2(e^(3x+1))=10
Well, that's down to the OP, seeing as he's left it quite ambiguous. However, TSR isn't a place where we just 'give' answers to questions, in essence, the answer is the 'full' solution, it leaves nothing for the OP to do, which is precisely what we don't want. Don't be a smart arse.
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TenOfThem
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(Original post by Phichi)
Well, that's down to the OP, seeing as he's left it quite ambiguous. However, TSR isn't a place where we just 'give' answers to questions, in essence, the answer is the 'full' solution, it leads nothing for the OP to do, which is precisely what we don't want. Don't be a smart arse.
As usual

PRSOM
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returnace
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#8
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2:2(e^3x+1)=10:2 we have divided the equation with 2 cos we want to find inside()
e^3x+1=5 e^3x=4 using ln=loge function for 2 side of equation
lne^3x=ln4 3x=ln4 x=ln4^3
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returnace
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hi
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TenOfThem
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#10
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(Original post by returnace)
x=ln4^3
Posting full solutions is against the rules, even when they are incorrect
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returnace
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#11
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Auhh so sorry,thx for notice,just tried to help
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