Assume that a population size at time t is N(t) and that N(t)=40 x 2^t , t>0.
Show that N(t)=40e^t ln2
Can someone please guide me step by step on this question? Im really confused. thanks.
I tried to work it out, and this is what I got.
N(t)= 40 x 2^t
= 40 x ln 2^t
= 40 x t ln 2
changing the base of a log: Alg II
http://www.sosmath.com/algebra/logs/log4/log43/log43.html
I'm sorry, I still don't understand
by definition, 2 = e^(ln 2)
Therefore, 2^t = e^(ln 2)^t = e^(t ln 2)
Make sense yet?
Now, ln (2^t) = t * ln 2
So, ln N = ln 40 + t ln 2
To show that N(t) = 40e^t ln2, we can use the properties of exponential and logarithmic functions. Here's a step-by-step guide:
1. Given N(t) = 40 x 2^t, we want to rewrite it in the form 40e^t ln2.
2. Start by expressing 2^t as e^(t ln 2). The reason for this is that the natural base of the exponential function e is related to the logarithm function ln.
3. We know that ln(x^a) = a ln(x) using the logarithmic property. Therefore, we can rewrite 2^t as e^(t ln 2).
4. Substitute the expression into N(t):
N(t) = 40 x e^(t ln 2).
5. Rearrange the factors:
N(t) = 40e^t ln2.
And that's the desired result: N(t) = 40e^t ln2.