If #3000# dollars invested in a bank account for #8# years, compounded quarterly, amounts to #4571.44# dollars, what is the interest rate paid by the account?

We'll use the compounding interest formula, #A=P(1+r/n)^(nt)#.
Here, #A# is the final amount; #P# is the principal, or original amount invested; #r# is the interest rate written in decimal form; #n# is the number of times the money is compounded per time #t#. Usually, #t# is in years and #n# is number of compoundings per year. Quarterly means #n=4#.

Plug in the information you have: #4,571.44=3,000(1+r/4)^(4*8)#
So #4,571.44=3,000(1+r/4)^32#
Divide both sides by #3,000# to get #1.52381=(1+r/4)^(32)#

Now, there are a couple of ways to solve this, with either roots or logarithms. Since you're in pre-calculus, I'll assume your teacher wants you to use logs.

Take the #log# of both sides: #log 1.52381=log (1+r/4)^(32)#
Remember the property of logarithms that says #log a^b=b log a#.
Bring the #32# in front of the expression on the right:
#log 1.52381=32log (1+r/4)#

Then divide both sides by #32#: #(log 1.52381)/32=log (1+r/4)#

Simplify: #.0057166=log (1+r/4)#

Using the inverse property of logarithms, change the expression to read: #10^.0057166 = 1+r/4#

Simplify the left: #1.02325 = 1+r/4#

Solve for #r# by subtracting #1# and multiplying by #4#.

#r=.053000#, or in percent notation, #r=5.3%#.