1290=160e(in2/2)(t)? hot do i caculate t?
what is 'in' ?? or is your question, \(\huge 1290=160e^{\ln 2/2}\)
yes thats right.
where does the t go ?
okk...then use \(\huge x^{b/2}= \sqrt {x^b}\) and \(\huge e^{ln x}=x \)
oh, yeah, what about t ? :P
t is also exponent next In2/2
and i have to solve for t
the first step is divide both sides by 160
like \(\huge 1290=160e^{t\ln 2/2}\) ?
no t in the end
anyways do what @phi said...
8.0625?
do you mean: \[ 1290= 160 e^{\frac{\ln 2}{2} t } \] that is the same as \[ 1290= 160 e^{\frac{t \ln 2}{2} } \]
oh i didnt know that! cool :)
next step is to take natural log on both sides...
how do i do that
I would do \[ \frac{1290}{160} = e^{t \frac{\ln 2}{2}} \] then write ln on both sides: \[\ln\left( \frac{1290}{160}\right) = \ln\left(e^{t \frac{\ln 2}{2}} \right)\]
ln "undoes" the e to the exponent. it brings down the exponent of e
in other words \[\ln(e^x)= x \]
no matter how complicated "x" is
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
Ok, ! i got 2.794
mmm... i entered the answer on online homework and it says it was wrong
Did you check your answer ? I did not get 2.794
i keep geting 2.794249
so far we have \[ \ln\left( \frac{1290}{160}\right) = t \frac{\ln 2}{2} \]
yes.
solve for t to get \[t= \frac{2}{\ln 2} \cdot \ln\left( \frac{1290}{160}\right) \]
what steps are you taking ?
oh i didnt know do In (1290/160) I just caculated without In
go back and look how we did it. We had to take the log of both sides.
i mean i didnt do. haha
I got 6.0224!
yes, that looks good. If you use that number in the original equation it will work out.
Thank you very much :) so excited when I got the number! you are the star!
thank you
Join our real-time social learning platform and learn together with your friends!