Question 735920
You are assuming that that compound interest formula is a good choice to use for this price growth, or the instruction given to you are to use it.  You could assume the compounding period is the individual year, so n=1.    t will be in years.  You'll use {{{A=P(1+r)^t}}}.  


Solve for r symbolically, and THEN plug in the values.  Use either common logs OR natural logs, but be consistent.  


I'm using natural logs for text convenience.
{{{ln(A)=ln(P(1+r)^t)}}}
{{{ln(A)=ln(P)+ln((1+r)^t)}}}
={{{ln(P)+t*ln(1+r)}}}
-----------
{{{ln(A)-ln(P)=t*ln(1+r)}}}
{{{(ln(A)-ln(P))/t=ln(1+r)}}}
You may want to rewrite in exponential form...
{{{highlight((1+r)=e^((ln(A)-ln(P))/t))}}}
Adjusting for the sum 1+r should not seem to be inconvenient.
[In case the top of the exponent is unreadable, click on "Show Source".]


Substitute the values and compute r or 1+r, using:
A=4.59
P=0.49
t=49