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Mathematics 5 Online
OpenStudy (vheah):

how would u solve for t if t is on the exponent??

OpenStudy (vheah):

\[0.1 = 0.5e ^{-250t}\]

jhonyy9 (jhonyy9):

make ln. on both sides so what will get ?

jhonyy9 (jhonyy9):

no sorry first divide both sides by 0,5 so what will get ?

jhonyy9 (jhonyy9):

0,1 = 0,5e^-250t divide both sides by 0,5 and will get 0,1/0,5 = e^-250t but you know that e^-250t = 1/(e^250t) yes ?

jhonyy9 (jhonyy9):

0,1/0,5 = 0,2 0,2 = 1/e^250t multiply both sides by e^250t and will get 0,2*e^250t = 1 divide both sides by 0,2 e^250t = 1/0,2 e^250t = 5 make ln on both sides ln e^250t = ln5 250t ln e = ln5 250t = ln5 t = ln5/250

jhonyy9 (jhonyy9):

@Nnesha your opinion please ? ty.

Nnesha (nnesha):

why are we multiplying by `e^{-250t} `

jhonyy9 (jhonyy9):

not by -250t just by 250t bc. there is 250t in denominator

Nnesha (nnesha):

jeez i lost all the work ohh my lord gzzzz one misssssclick...

jhonyy9 (jhonyy9):

e^-250t = 1/e^250t yes bc. there is exponent with minus in front

Nnesha (nnesha):

\[\large\rm 0.1 =0.5 e^{-250t}\] like jhonny mentioned divide both sides by .5 \[\large\rm \frac{0.1}{0.5}=\frac{0.5}{0.5}e^{-250t}\] \[\large\rm 0.2 = e^{-250t}\] now take ln both sides

Nnesha (nnesha):

no we don't need to do that once we take ln e would cancel out left with ln(0.2)=-250t

Nnesha (nnesha):

and then simply solve for t :=))

jhonyy9 (jhonyy9):

so this mean that my above wrote work not is right ?

Nnesha (nnesha):

yes i think so i believe that step is not necessary we should not be care about negative exponent because we know the fact ln and e cancel each otherz out

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