1. ## Logarithmic Applications

I'm doing a compound interest problem and I wanted to know if I am doing it correctly. The problem is

How long will it take for $1,000 to grow to$5,000 at an interest rate of 3.5% if interest is compounded

a) quarterly

b) continuously

For part a) I used the formula A = P(1 + r/n)^nt

5,000 = 1,000( 1 + .035/4)^4t

5 = (1.00875)^4t

ln5 = 4t ln1.00875

ln5 / 4ln1.00875 = T

Did I do it correctly? Perhaps i'm not putting it into the graphing calculator correctly but I keep getting the wrong answer compared to the answer section of the book. If I did do it correctly, can someone let me know what is the correct way to punch that into the calculator? I have a a Ti-83 graphing calculator.

For part b) I used the continuous forumula of A = Pe^rt

5,000 = 1,000e^.035t

5 = e^.035t

ln5 = .035t

ln5 / .035 = t

Did I do that correctly?

Thanks for any help

2. Hello, Jonathan!

How long will it take for $1,000 to grow to$5,000
at an interest rate of 3.5% if interest is compounded

a) quarterly

b) continuously

For part a) I used the formula: . $A \;= \;P\left(1 + \frac{r}{n}\right)^{nt}$

Then: . $5,000 \;= \;1,000\left(1 + \frac{0.035}{4}\right)^{4t}\quad\Rightarrow\quad5 \;= \;(1.00875)^{4t}$

. . $\ln5 \:= \:4t\!\cdot\!\ln(1.00875)\quad\Rightarrow\quad t \:=\:\frac{\ln5}{4\!\cdot\!\ln(1.00875)}$
You are probably entering it incorrectly on your calculator.
Here's one common error . . .

We want: . $(\ln 5) \div \left[4\!\cdot\!\ln(1.0875)\right]$

We must enter: . $\boxed{\ln}\;\boxed{5}\;\boxed{)}\;\boxed{\div}\;\ boxed{(}\;\boxed{4}\;\boxed{\times}\;\boxed{\ln}\; \boxed{1.0875}\;\boxed{)}\;\boxed{=}$
. . . . . . . . . . . . . . . . . ${\color{red}\uparrow}$
When we press $\boxed{\ln}$, the display shows: . $\boxed{\ln\,(\qquad}$
. . It automatically supplies the left parenthesis.

After we have entered the $\boxed{5}$, we must close the parentheses with $\boxed{)}$

If we don't, we will have: . $\boxed{\ln}\;\boxed{5}\;\boxed{\div}\;\boxed{4} \cdots$
. . which the calculator interprets as: . $\ln\left(\frac{5}{4}\right)$