Question

Medal +fan +testimonial for help with my last two questions for homework :)

Answer 1

HI!!

Answer 2

Hiya @misty1212 :)

Answer 3

you know how to solve an exponential equation? where the variable is in the exponent?

Answer 4

uhmm sorta i have this great mate called mathway but thats about it ;)

Answer 5

first off we need to set up the equation with the actual numbers

Answer 6

ok great

Answer 7

you want the half life, and \(k=0.00011\) so set \[\huge e^{0.00011t}=0.5\] and solve for \(t\) in two steps

Answer 8

ok im with you

Answer 9

first write in logarithmic form as
\[\huge 0.00011t=\ln(0.5)\] then divide to find \(t\)

Answer 10

ok so i got t=-6301.34? doesnt sound right

Answer 11

ok i made a mistake and dropped the minus sign for some reason

Answer 12

it is right except i forgot the minus sign
you got \(-6301\) but it should be \(6301\)

Answer 13

ok so 6301

Answer 14

so the answer for 5 is 6301 years?

Answer 15

yeah it should have been \[\huge e^{-0.00011t}=0.5\] etc

Answer 16

rounded to the nearest year, yes

Answer 17

ok great thank you for your help :)

Answer 18

could you also help with the second one please?

Answer 19

yeas sure it is the same, only this time you want to know when it decays to one third of its original amount

Answer 20

if you start with 27 and end up with 9, it is one third
if you need convincing , set
\[27e^{-.00011t}=9\] and the first step to solve is divide by \(27\) and get \[e^{-0.00011t}=\frac{1}{3}\]

Answer 21

ok so what numbers do i have to divide by 27?

Answer 22

solve just like the last one only with \(\frac{1}{3}\) instead of \(\frac{1}{2}\)

Answer 23

you got that?

Answer 24

not really :/

Answer 25

i got 9900 years?