Find the exponential function y=Ce^kt that passes through the two points: (2,3/2) and (4,5)

Question

amistre64 · Answer

we know t and y, should we solve for k?

anonymous · Answer

K and C

amistre64 · Answer

$$5=Ce^{4k}$$
$$\frac{3}{2}=Ce^{2k}$$
hmmm

anonymous · Answer

Correct.

amistre64 · Answer

$$ln(5)-ln(C)=4k$$
$$ln(3/2)-ln(C)=2k$$

$$ln(5)-ln(C)=4k$$$$-2ln(3/2)+2ln(C)=-4k$$

$$ln(5)-ln(C)-2ln(3/2)+2ln(C)=0$$

$$ln(C)=ln(9/4)-ln(5)$$

$$ln(C)=ln(9/20)$$

$$C=9/20$$
maybe?

anonymous · Answer

@amistre64, Can I try?

amistre64 · Answer

$$ln(5)-ln(9/20)=4k$$
$$ln(100/9)=4k$$
$$\frac{1}{4}ln(100/9)=k$$

its prolly wrong, but we can test it out at least :)

amistre64 · Answer

yep, i was never good at the exponential finding stuff to well ;)

anonymous · Answer

What I got was:

$$k = \ln (10/3) \times (1/2) $$

and C = .45

but I am not sure those are correct.

anonymous · Answer

Divide to get K:  e^2k = 10/3
2k = ln ( 10/3)
->k = (1/2) ln ( 10/3)

amistre64 · Answer

http://www.wolframalpha.com/input/?i=y%3D%289%2F20%29+e%5E%28ln%28100%2F9%292%2F4%29 i think the wolf likes it

anonymous · Answer

Then plug it back:
C = 5/ e^( 2ln10/3)

amistre64 · Answer

$$\large y=\frac{9}{20}e^{\frac{1}{4}ln(100/9)t}$$
works for me ... but like i said, i gots no idea if its what your looking for

anonymous · Answer

My PC and connection, including me are so slow! so pls do ask me if you need more explanation!

amistre64 · Answer

9/20 = .45 so we agree on that http://www.wolframalpha.com/input/?i=ln%28100%2F9%29%2F4 says the 1/2 ln(10/3) is the same

anonymous · Answer

I just checked. $$y=.45e ^{\ln (10/3)\times(1/2)t} $$ works with the points

amistre64 · Answer

yep, it does :)

anonymous · Answer

Sweet! Thanks for helping!