Landon opened a savings account with a deposit of $3,786.45. Nine years later, the balance in the account is $5,135.29. If the interest rate is 3.4%, how often does the interest compound?

Question

anonymous · Answer

$$A=P\left(1+{r\over100}\right)^n$$
A=5135.29
P=3768.45
r=3.4
n=?
plug in and solve.

anonymous · Answer

ooh.. they want "how often"... k

anonymous · Answer

my bad.. n = 9
and we use
$$A=P\left(1+{r\over 100k}\right)^{kn}$$

anonymous · Answer

what would i use for k ?

anonymous · Answer

$$5135.29=3768.45\left(1+{3.4\over100k}\right)^{9k}\
{5135.29\over3768.45}=\left(1+{0.034\over k}\right)^{9k}\
1.3627=\left(1+{0.034\over k}\right)^{9k}\
\log_{10}(1.3627)=9k\times\log_{10}\left(1+{0.034\over k}\right)\
0.1344=9k\times\log_{10}\left(1+{0.034\over k}\right)
$$

anonymous · Answer

the only way I see is by trial and error

1) usually savings accounts are coumponded monthly..
so, check for     k = 12
2) if does not work, check semiannually        k=2
3) check annually.         k=1

anonymous · Answer

alright thanks

anonymous · Answer

the answer must be "ANNUALLY" , with k=1