Question

The function f(x) = 15(2)^x represents the growth of a frog population every year in a remote swamp. Elizabeth wants to manipulate the formula to an equivalent form that calculates every half-year, not every year. Which function is correct for Elizabeth's purposes?
f(x) = 15(2^1/2)^2x
f(x) = 15(2^2) ^x/2
f(x) = 15/2(2)^x
f(x) = 30(2)^x

Answer 1

Half one is 1/2;
half three is 3/2;
Thus, if f(x) represents annual growth, f(x)/2 shows it every half-year, like this:
$\dfrac{f(x)}{2} = \dfrac{15\cdot 2^x}{2} \\ \\ \dfrac{f(x)}{2} = \dfrac{15}{2}\cdot 2^x$

Written in plain text: f(x)=15/2*2^x

Answer 2

Answer:

$\boxed{\boxed{B.\ f(x)=15(2^2)^{\frac{x}{2}}}}$

Step-by-step explanation:

The general equation for exponential growth is,

$y=a(1+r)^x$

where,

a = initial amount,

r = rate of growth,

y = future amount,

x = time.

The function $f(x) = 15(2)^x$ represents the growth of a frog population every year in a remote swamp.

where 15 is the initial amount of frog.

As Elizabeth wants to manipulate the formula to an equivalent form that calculates every half-year, so putting $x=\dfrac{x}{2}$ in the general function,

$y=15(1+r)^{\frac{x}{2}}$

Now we have to find the value of r.

As we will get same future amount, irrespective to the function used, so comparing the old function with the new function,

$\Rightarrow 15(1+r)^{\frac{x}{2}}=15(2)^x$

Multiplying the exponents of both sides by $\dfrac{1}{x}$

$\Rightarrow (1+r)^{\frac{x}{2}\times \frac{1}{x}}=(2)^{x\times \frac{1}{x}}$

$\Rightarrow (1+r)^{\frac{1}{2}}=(2)^1$

Squaring both sides,

$\Rightarrow (1+r)=(2)^2=4$

$\Rightarrow r=4-1=3$

Putting the value of r, in the general equation,

$y=15(1+3)^{\frac{x}{2}}=15(4)^{\frac{x}{2}}=15(2^2)^{\frac{x}{2}}$