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:[tex]\boxed{\boxed{B.\ f(x)=15(2^2)^{\frac{x}{2}}}}[/tex]Step-by-step explanation:The general equation for exponential growth is,[tex]y=a(1+r)^x[/tex]where,a = initial amount,r = rate of growth,y = future amount,x = time.The function [tex]f(x) = 15(2)^x[/tex] 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 [tex]x=\dfrac{x}{2}[/tex] in the general function,[tex]y=15(1+r)^{\frac{x}{2}}[/tex]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,[tex]\Rightarrow 15(1+r)^{\frac{x}{2}}=15(2)^x[/tex]Multiplying the exponents of both sides by [tex]\dfrac{1}{x}[/tex][tex]\Rightarrow (1+r)^{\frac{x}{2}\times \frac{1}{x}}=(2)^{x\times \frac{1}{x}}[/tex][tex]\Rightarrow (1+r)^{\frac{1}{2}}=(2)^1[/tex]Squaring both sides,[tex]\Rightarrow (1+r)=(2)^2=4[/tex][tex]\Rightarrow r=4-1=3[/tex]Putting the value of r, in the general equation,[tex]y=15(1+3)^{\frac{x}{2}}=15(4)^{\frac{x}{2}}=15(2^2)^{\frac{x}{2}}[/tex]