Lions were introduced in the park in 2010. In 2012, the park has a population of 32 lions. A year later, in 2013, the park has 128 lions. Write an exponential function that models this situation. Let x represent the number of years since 2010 and let f(x)represent the number of lions.

You can use the fact that the given figures are expressible as 2 raised to some powers.

The exponential function modeling the number of lions to the years passed since 2010 is given as

Those functions which are expressed as some constant raised to the power of the argument called exponential function.

Since 2012  - 2010 = 2 and the number of lions = 32 =

and 2013 -2010 = 3 and the number of lions =128 =

Thus, there is increment of 2 exponent as 1 year was changed.

Thus, we have:

where a might be some constant( i took this form of exponent just by trial and error. It was already visible that one year was increasing double output, thus 2x and since there was just double increase and no other effect, there might be some non variable thing(a constant) so i thought to try some variable and see if this can work)

Since at x = 2, we get 2 raised to the power 5

thus.

Thus,

The exponential function modeling the number of lions to the years passed since 2010 is given as

You can see that this will give correct output for x = 2 and x = 3

Learn more about exponential functions here:

link

y = 2(4^(x))

Step-by-step explanation:

In 2012, the park has a population of 32 lions. A year later, in 2013, the park has 128 lions.

This means in 2012, it was 2^(5) lions while after 1 year it was 2^(7) lions

Thus, it increases every year by a ratio of 2^(2)

Thus, in 2011, it was 2^(5) - 2^(2) = 2^(3) lions

In 2010, it was 2^(3) - 2^(2) = 2

Since Lions were introduced in the park in 2010, using f(x), it means;

at x = 0, f(x) = 2

at x = 1, f(x) = 2^(3)

at x = 2, f(x) = 2^(5)

at x = 3, f(x) = 2^(7)

To find an exponential function that models the situation, let's use the generic formula: y = ab^(x)

At x = 0 and y = 2, we have;

2 = a × b^(0)

Thus,a = 2

At x = 1, y = 2^(3)

Thus; 2^(3) = 2 × b^(1)

b = 8/2

b = 4

So the exponential equation is;

y = 2(4^(x))

are ya tryin to ask something or is this a statement