Question

Mary deposited $350 in a bank account that promises 2.8 percent interest compounded continuously. Approximately how many years will it take to reach a balance of $500?

1.43 years

2.80 years

5.55 years

12.77 years

Answer 1

come back @saifoo.khan

Answer 2

Lets try again bro

Answer 3

y = years
350 x (350 x 2.8)y =500

Answer 4

SORRY
I meant this
350 + (350 x 2.8)y =500

Answer 5

Continuous interest tells you that you will use the interest formula that involves e.
A=Pert

A is the future value. P is the starting value. r is the rate as a decimal. t is the number of years of the investment.
You will need to use ln to get the exponent to "come down" so the equation becomes easier to manage.

Answer 6

yea you right but that answer didnt tell me the years lol

Answer 7

figure it out breh

Answer 8

@phi

Answer 9

@Preetha

Answer 10

@jdoe0001

Answer 11

hey

Answer 12

can yall help?

Answer 13

have you covered logarithms yet?

Answer 14

Thats what im on

Answer 15

k

Answer 16

lol

Answer 17

Mary deposited $350 in a bank account that promises 2.8 percent interest compounded continuously. Approximately how many years will it take to reach a balance of $500?

$$ \Large { A = Pe^{rt}
\\ \therefore \\
A = 350 e^{.028t }
\\~\\ \text{when does A reach 500?}\\\therefore\\
\\500 = 350 e^{.028t }

}
$$

Answer 18

hmmm actauly.... lemme fix that... the term inside is a two factor

Answer 19

but you'd end up with a log function anyhow
you;'d take the log to both sides first, and then solve for "t" in the end

Answer 20

ok

Answer 21

I got a 5 5 5

Answer 22

555?

Answer 23

yea

Answer 24

Im prolly wrong

Answer 25

\(\bf \qquad \textit{Compound Continuously Interest Amount}\\
A=Pe^{rt}
\qquad
\begin{cases}
A=\textit{current amount}\to &500\\
P=\textit{original amount deposited}\to &\$350\\
r=rate\to 2.8\%\to \frac{2.8}{100}\to &0.028\\

t=years
\end{cases}
\\ \quad \\
A=Pe^{rt}\implies ln(A)=ln(Pe^{rt})\implies
ln(A)=ln(P)+ln(e^{rt})\)

Answer 26

so its 2.80

Answer 27

recall the \(\bf \textit{log cancellation rule of }log_{\color{brown}{ a}}{\color{brown}{ a}}^x=x\qquad thus
\\ \quad \\
ln(e^\square)\implies log_{\color{brown}{ e}}{\color{brown}{ e}}^\square \implies \square \)

Answer 28

hehe

Answer 29

maybe its 1.43

Answer 30

am I wrong

Answer 31

hmmm nope... so... solve it for "t", see what the equation gives you
then put in the values given for the current balance, the principal and the rate

Answer 32

ok

Answer 33

you can come back

Answer 34

I got 12 years

Answer 35

giving me a headache

Answer 36

what's "t" equal to in the equation though?

Answer 37

t= years

Answer 38

I know t=years

Answer 39

Im plugging it in

Answer 40

maybe I have the wrong calculator

Answer 41

500=350e^0.028(t)

Answer 42

350e^0,028(12.77)

Answer 43

\[350e^0.028(12.77)\]

Answer 44

well. in "t" is not quite in "e" terms though

Answer 45

so you want me to put years for e?

Answer 46

this sht here

Answer 47

hmmm nope... you're meant to solve for "t"

Answer 48

and then you can just replace the values for "r" and "P" and "A" for the given amounts for rate, principal and amount,, respectively

Answer 49

t = (1/r)(A/P - 1)

Answer 50

(1/0.028)(500/350-1)

Answer 51

I dont know Im done

Answer 52

well it is 12.77