Question

f(t)=Q0(1+r)^t. Find the growth rate, r, to the nearest thousandth, given f(0.01)=1.06 and f(0.11)=1.09.

Answer 1

well substituting you have

\[1.06 = Q_{0}( 1 + r)^{0.01}\]

and you also have

\[1.09 = Q_{0}(1 + r)^{0.11}\]

using a little index law property about adding powers when multiplying you can rewrite the 2nd equation as

\[1.09 = Q(1 + r)^{0.01} \times (1 + r)^{0.1}\]

now you can do a little substitution from the 1st equation

\[1.09 = 1.06 \times (1 + r)^{0.1}\]

divide both sides of the equation by 1.06

\[\frac{1.09}{1.06} = (1 + r)^{0.1}\]

since 0.1 = 1/10 you have an equation where

\[\frac{1.09}{1.06} = ( 1 + r)^{\frac{1}{10}}\]

raise each term to the 10th power, then

\[(\frac{1.09}{1.06})^{10} = 1 + r\]

this should make it easy to find the growth rate r...