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calc II separable equation, please help :)

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\[\frac{ du }{ dt }=e ^{6u+8t}\] given that u(0)=8. So this is how I set it up \[\int\limits_{}^{}e ^{-6u}du=\int\limits_{}^{}e ^{8t}dt\]
Is that right so far?
yes looks right. so far.

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Okay, so this is what I got for my integration: \[\frac{ -e ^{-6u} }{ 6 }=\frac{ e ^{8t} }{ 8 }+c\]
and then using that u(0)=8, I found that c=-0.125.
I actually have already worked through this problem, but I got the wrong answer at the end so I'm trying to figure out where I'm going wrong
so then I end up with \[\frac{ -e ^{-6u} }{ 6 }=\frac{ e ^{8t} }{ 8 }-0.125\]
so then my final answer was u(t)=\[\frac{ 1 }{ 6 }\ln(\frac{ 3 }{ 4 }e ^{8t}-\frac{ 3 }{ 4 })\]
but the book says that isn't the right answer
hmm well i also got c same as u.. well what answer book says?
well, it's actually webwork, so it just tells me that my answer is wrong, I'm not sure what the right answer is
I'm thinking something must be wrong with the way I'm doing the algebra to rearrange it in terms of u(t)?
well i am not sure but the way u solved question looks fine and the answer should be the one u got..
hmmm. Okay. I'll ask the tutor at 8. Thanks anyway :)

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