Answer/Step-by-step explanation:
The exponential function modeling the number of lions ⇒ the years passed since 2010 is appointed as
[tex]f(x)=2^1^+^2^x[/tex]
Well, If you have an question like what is exponential function?
Let's me tell you:
exponential functions which are expressed as some constant raised to the power of the argument called exponential function.
You might be asking how to find which function can model the given question..
[tex]\mathrm{Since \;2012 - 2010 = 2 \;and\; the\; number \;of\;lions = 32 =2^5}[/tex]
[tex]\mathrm{and \;2013 -2010 = 3\; and\; the\; number \;of\; lions\; =128 = 2^7}[/tex]
Hence, there is increment of 2 exponent as 1 year was changed.
Now, we have:
[tex]f(x)=2^2^x^+^a[/tex]
[tex]\mathrm{Thus,\; it\; increases \;every \;year \;by\; a \;ratio \;of \;2^(^2^)}[/tex]
[tex]\mathrm{Thus, \;in \;2011, \;it \;was \;2^(^5^) - 2^(^2^) = 2^(^3^)\; lions}[/tex]
[tex]\mathrm{In\; 2010, it\; was 2^(^3^) - 2^(^2^) = 2}[/tex]
Since Lions were introduced in the park in 2010, using f(x), it means;
[tex]x = 0, f(x) = 2[/tex]
[tex]x = 1, f(x) = 2^(^3^)[/tex]
[tex]x = 2, f(x) = 2^(^5^)[/tex]
[tex]x = 3, f(x) = 2^(^7^)[/tex]
Let's use the generic formula: y = ab^(x)
At x = 0 and y = 2, we have:
[tex]2 = a * b^(^0^)[/tex]
Thus,a = 2
At x = 1, y = 2^3
b = 8/2
b = 4
So the exponential equation is;
[tex]y = 2(4^(^x^))[/tex]
~Learn with Lenvy~
