Question 130·Medium·Advanced Math

In function notation, $h(2)=12$ means input $2$ gives output $12$. Substitute both given inputs for $x$ in the rule:

The exponents are $0$ and $2$, respectively. The $+7$ stays outside each power, so dividing $27$ by $12$ would not give the base.

Type $\left[kb^{2-2}+7,kb^{4-2}+7\right]\sim\left[12,27\right]\left\{b>0\right\}$. This regression fits the two equations together; $\sim$ tells Desmos to find the constants, and the restriction keeps the positive base the question requires. Under PARAMETERS, Desmos shows $k=5$ and $b=2$.

Type $h(x)=kb^{x-2}+7$ below the regression. Desmos uses the fitted constants from the line above, so you've defined $h(x)=5(2)^{x-2}+7$. Keep the $+7$ outside the power.

Type $h(5)$. Desmos prints $47$: putting $5$ into the rule makes the exponent $5-2=3$. So the value of $h(5)$ is $47$. Choice C.

The $+7$ is a fixed shift, so the exponential part is $h(x)-7$. Find that part at each given input:

These are the amounts the base multiplies, unlike the full outputs $12$ and $27$.

The base $b$ multiplies the amount above $7$ each time $x$ increases by $1$. Going from $2$ to $4$ uses it twice, so compare the two amounts:

Divide:

The problem says $b$ is positive, so take the positive square root of $b^2=4$:

The negative square root breaks the given condition.

From $x=4$ to $x=5$ is one step, so multiply the amount above $7$, which is $20$, by $2$:

Add the fixed $7$:

So the value of $h(5)$ is $47$. Choice C.