radioactive life

[Guest]

$3500 are invested in an account that pays interest of the annual rate of 4%. Determine equation that gives balance in the amount t years after inception of account. f(t)=3500(1.04)t

Half-life of radioactive substance is 300 years. Fill in the following table if initial amount is 900 lb.

t f(t)
0 900
300 450
600 225

This is what I got so far:

ab ^0=900 ----> a=900
ab ^300=450 ----> 900b ^300/900=450/900
b ^300=.5 x 1/300

Am I on the right track?

 

[soroban]

Hello, Angela!

In the half-life problem, you were fine up to the last step . . .

Half-life of radioactive substance is 300 years.

Fill in the following table if initial amount is 900 lb.

. t . . . . .f(t)
. 0 . . . . 900
300 . . . 450
600 . . . 225

This is what I got so far:

$\displaystyle a\cdot b ^0\,=\,900\;\;\Rightarrow\;\;a\,=\,900$

$\displaystyle 900\cdot b^{300}\,=\,450\;\;\Rightarrow\;\;b^{300}\,=\,0.5$

Last step: \(\displaystyle \;b\:=\0.5)^{\frac{1}{300}}\:\approx\:0.99769\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Are you supposed be using "e" in this problem?

Then: $\displaystyle \;f(t)\:=\:a\cdot e^{-kt}$

Since $\displaystyle f(0)\,=\,900:\;\;a\cdot e^0\,=\,900\;\;\Rightarrow\;\;a\,=\,900$

. . and we have: $\displaystyle \;f(t)\:=\:900\cdot e^{-kt}$


Since $\displaystyle f(300)\,=\,40:\;\;900\cdot e^{-300k}\,=\,450\;\;\Rightarrow\;\;e^{-300k}\,=\,0.5$

. . -$\displaystyle 300k\,=\,\ln(0.5)\;\;\Rightarrow\;\;k\,=\,\frac{\ln(0.5)}{-300}\:\approx\;0.00231$

Therefore: $\displaystyle \;f(t)\:=\:900\cdot e^{-0.00231t}$

 

[Guest]

How do you do that on the calculator, because I got .00166 when I did (.5) x (1/300).

 

[soroban]

Hello, angelasloan7038!

How do you do that on the calculator, because I got .00166 when I did (.5) x (1/300).

You're multiplying!

$\displaystyle \text{It says: }(0.5)^{\frac{1}{300}}\;\cdots\; \text{That's }0.5\text{ to the }\frac{1}{300}\;power!$



Do you know that $\displaystyle 7^2$ does not mean $\displaystyle 7\,\times\,2$ ?

. . . . . . . . . . .that $\displaystyle 3^{\frac{1}{2}}$ does not mean $\displaystyle 3\,\times\,\frac{1}{2}$ ?