An employee's salary is calculated by the formula S(t)=S0 e^.08t. t=0 corresponds to the beginning of 2011. in 2014 this employee was paid $69,918.70. What was the employee's annual percent salary increase?

Question

tkhunny · Answer

How many years have passed?

anonymous · Answer

t=0 at the beginning of 2011. The salary $69,918.70 is in the year 2014, so 3 years.

tkhunny · Answer

All right.  What, then is $A_{0}$?

anonymous · Answer

That is what I'm trying to solve, isn't it? When I went through what I thought was the process, I got an answer of 46. something, which is definitely not the answer.

tkhunny · Answer

Of course, I meant $S_{0}$.

You have: $S(t)=S_{0}e^{.08t}$
You have: $t = 3$
You have: $S(3) = 69,918.70$

Solve for $S_{0}$

tkhunny · Answer

$69918.70 = S_{0}e^{0.08(3)}$ -- Solve for $S_{0}$.

anonymous · Answer

I must be doing something wrong. Here's my process:
$$S(t)=S _{0}e ^{0.08t}$$
$$69918.7=S _{0}e ^{0.08(3)}$$
$$69918.7=S _{0}e ^{.24}$$
$$\ln 69918.7=S _{0}(.24)\ln e$$
with the natural log canceling out e on the right side
$$11.155=S _{0}(.24)$$
Divide both sides by .24 and I get
$$S _{0}=46.4795$$

anonymous · Answer

however, if I don't use natural log and just divide by e^.24, I get S0= 54999.99. Is this correct? If so, why would I not use natural log?

tkhunny · Answer

Try: 

$\ln(69918.70) = \ln(S_{0}) + 0.24$

How did $S_{0}$ manage to avoid the logarithm?