00:01
Now this problem we have been given these four points, and it is suspected that they lie on the graph of y equals c times 2 to the tx for constant c and t, and we want to determine if this is true or not.
00:14
Now, in order to do this, let's take these first two points, use them to find the values of c and t, and then we'll go from there.
00:22
And so let's put in for that first point, zero for x, negative point three for y.
00:29
So they have 0 .3 is equal to c times 2 to the 0 power.
00:35
And so that tells us c is negative 0 .0 .5.
00:41
Now for that second one, we have negative 0 .345 as equal to negative 0 .3 times 2 to the t power.
00:55
Again, all we did was plug in negative 0 .35 for y and negative point, i'm sorry, and 1 for x.
01:02
And then remember we just found that c was negative 0 .3.
01:06
And then now we can solve this for t.
01:09
Remember we solve it for t...