Question

can anyone teach me how to solve this?!...

Answer 1

Zoey opened a savings account four years ago with a deposit of $935.15. The interest on the account compounds monthly. If Zoey's current account balance is $1,128.16, what is the interest rate on the account?

Answer 2

\[A=P(1+\frac{r}{n})^{n\cdot t}\]

Answer 3

i know the formula but i still cannot solve it, \[193.01= (1 + (r/12))^{48}\]

Answer 4

but then what?

Answer 5

Use the logarithm function to bring down the exponent.

Answer 6

and how would i do that?

Answer 7

\[ \log(193.01)=\log\left(1+\frac{r}{12}\right)^{48}\]
\[ \log(193.01)=48\log\left(1+\frac{r}{12}\right)\]

Answer 8

how would i continue solving this?

Answer 9

like what do i do with the 12...

Answer 10

\[\frac{\log(193.10)}{48}=\log\left(1+\frac{r}{12}\right)\]

\[10^{\frac{\log(193.10)}{48}}=10^{\log\left(1+\frac{r}{12}\right)}\]

\[10^{\frac{\log(193.10)}{48}}=1+\frac{r}{12}\]

Answer 11

what is that 10?

Answer 12

The base for the log function \[ \log_{10}\].

Also the left had side should be 1.20639 and not 193

Answer 13

so, it will be .0476= log(1+(r/12))?

Answer 14

Where did the 0.476 come from?

Answer 15

om guessing log(193.01)=2.286/48=.0476

Answer 16

OK, where did the 193.01 come from?

Answer 17

A minus P... 1,128.16-935.15=193.01?

Answer 18

Try \(\dfrac{A}{P}\) instead.

Answer 19

.1370?

Answer 20

then, how do you get 1.20639 ?

Answer 21

\[ A=P\left(1+\frac{r}{n}\right)^{nt}\]
\[ \frac{A}{P}=\left(1+\frac{r}{n}\right)^{nt}\]
\[ \log\frac{A}{P}=(nt)\log\left(1+\frac{r}{n}\right)\]
\[ \frac{\log\frac{A}{P}}{nt}=\log\left(1+\frac{r}{n}\right)\]
\[ 10^{\frac{\log\frac{A}{P}}{nt}}=\left(1+\frac{r}{n}\right)\]
\[ 10^{\frac{\log\frac{A}{P}}{nt}}-1=\frac{r}{n}\]
\[ r=n(10^{\frac{\log\frac{A}{P}}{nt}}-1)\]

Answer 22

i still don't get it...
ok, the options are a).4%
b)4.7%
c)14.2%
d)18.9%
if you solve do you get any of this answers? cuz i dont

Answer 23

?