#### Answer

For $d=-2$,
There were 20 finches two decades before measurement.
For $d=0$,
There were $45$ finches at the time of measurement.
For $d=1$,
There will be about 63 finches a decade after the time of measurement.

#### Work Step by Step

For $d=-2$,
$ 45\cdot 1.5^{-2}=\qquad$ apply: $\displaystyle \mathrm{a}^{-n}=\frac{\mathrm{l}}{\mathrm{a}^{n}}$
$=45\displaystyle \cdot\frac{1}{1.5^{2}}=\qquad$ evaluate $1.5^{2}$
$=\displaystyle \frac{45}{2.25}\times\frac{4}{4}= \qquad$ eliminate the decimals in the denominator
$=\displaystyle \frac{45\times 4}{9}=5\times 4=20$
There were 20 finches two decades before measurement
For $d=0$,
$ 45\cdot 1.5^{0}=\qquad$ apply: $a^{0}=1$
$=45\cdot 1=45$
There were $45$ finches at the time of measurement
For $d=1$,
$45\cdot 1.5=62.5\approx 63$
There will be about 63 finches in a decade's time from the time of measurement