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Calculus1 19 Online
OpenStudy (anonymous):

1290=160e(in2/2)(t)? hot do i caculate t?

hartnn (hartnn):

what is 'in' ?? or is your question, \(\huge 1290=160e^{\ln 2/2}\)

OpenStudy (anonymous):

yes thats right.

OpenStudy (phi):

where does the t go ?

hartnn (hartnn):

okk...then use \(\huge x^{b/2}= \sqrt {x^b}\) and \(\huge e^{ln x}=x \)

hartnn (hartnn):

oh, yeah, what about t ? :P

OpenStudy (anonymous):

t is also exponent next In2/2

OpenStudy (anonymous):

and i have to solve for t

OpenStudy (phi):

the first step is divide both sides by 160

hartnn (hartnn):

like \(\huge 1290=160e^{t\ln 2/2}\) ?

OpenStudy (anonymous):

no t in the end

hartnn (hartnn):

anyways do what @phi said...

OpenStudy (anonymous):

8.0625?

OpenStudy (phi):

do you mean: \[ 1290= 160 e^{\frac{\ln 2}{2} t } \] that is the same as \[ 1290= 160 e^{\frac{t \ln 2}{2} } \]

OpenStudy (anonymous):

oh i didnt know that! cool :)

hartnn (hartnn):

next step is to take natural log on both sides...

OpenStudy (anonymous):

how do i do that

OpenStudy (phi):

I would do \[ \frac{1290}{160} = e^{t \frac{\ln 2}{2}} \] then write ln on both sides: \[\ln\left( \frac{1290}{160}\right) = \ln\left(e^{t \frac{\ln 2}{2}} \right)\]

OpenStudy (phi):

ln "undoes" the e to the exponent. it brings down the exponent of e

OpenStudy (phi):

in other words \[\ln(e^x)= x \]

OpenStudy (phi):

no matter how complicated "x" is

OpenStudy (phi):

the left side is some number the right side is \[ \frac{\ln 2}{2} t \] solve for t by multiplying both sides by 2/ln 2

OpenStudy (anonymous):

Ok, ! i got 2.794

OpenStudy (anonymous):

mmm... i entered the answer on online homework and it says it was wrong

OpenStudy (phi):

Did you check your answer ? I did not get 2.794

OpenStudy (anonymous):

i keep geting 2.794249

OpenStudy (phi):

so far we have \[ \ln\left( \frac{1290}{160}\right) = t \frac{\ln 2}{2} \]

OpenStudy (anonymous):

yes.

OpenStudy (phi):

solve for t to get \[t= \frac{2}{\ln 2} \cdot \ln\left( \frac{1290}{160}\right) \]

OpenStudy (phi):

what steps are you taking ?

OpenStudy (anonymous):

oh i didnt know do In (1290/160) I just caculated without In

OpenStudy (phi):

go back and look how we did it. We had to take the log of both sides.

OpenStudy (anonymous):

i mean i didnt do. haha

OpenStudy (anonymous):

I got 6.0224!

OpenStudy (phi):

yes, that looks good. If you use that number in the original equation it will work out.

OpenStudy (anonymous):

Thank you very much :) so excited when I got the number! you are the star!

OpenStudy (phi):

thank you

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