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

`\boxed{\boxed{B.\ f(x)=15(2^2)^{(x)/(2)}}}`

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=(x)/(2)` in the general function,

`y=15(1+r)^{(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)^{(x)/(2)}=15(2)^x`

Multiplying the exponents of both sides by
`(1)/(x)`

`\Rightarrow (1+r)^{(x)/(2)* (1)/(x)}=(2)^{x* (1)/(x)}`

`\Rightarrow (1+r)^{(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)^{(x)/(2)}=15(4)^{(x)/(2)}=15(2^2)^{(x)/(2)}`