The model gives population after $t$ years as $P(t)=12,000(1.03{)}^t$. To find when the population reaches $18,000$, set $P(t)$ equal to $18,000$:

This directly matches the situation, so that is the correct equation.

Shifting the graph of $y=2^x$ right 3 units replaces $x$ with $x-3$, giving $y=2^{x-3}$. Reflecting across the $x$-axis multiplies all output values by $-1$, so the new function is $y=-2^{x-3}$.

Since the models are equal at $t=1$, substitute $t=1$ into both expressions:

so

Squaring both sides gives

so

The statement that the models are equal at exactly one time is consistent with this value. With $k=8$, the equation becomes

At $t=1$, both sides equal $3$. Also, the exponential model increases much faster than the square-root model, so after meeting at $t=1$, they do not meet again. Thus the required value is $8$.

From the table,

Dividing gives

so

Then

which gives

So the model is

The condition

means

Now solve:

Multiply both sides by $4$:

Then

Therefore,

So the correct answer is $7$.

Let $a=2^x$ and $b=2^{-x}$. Then $ab=2^x\cdot 2^{-x}=1$, and the given equation becomes $a+b=\sqrt{10}$. We want $a^2+b^2$. Use the identity

Substituting the known values gives

So,

Because $a^2=2^{2x}$ and $b^2=2^{-2x}$, the value of $2^{2x}+2^{-2x}$ is $8$.