00:01
Alright, thank you for submitting this problem.
00:03
We are being asked in this problem to sketch y equals log base 2 of x as the inverse of an exponential.
00:10
Then we're going to be asked to use that graph to approximate some solutions here.
00:14
And then we want to solve and compare the estimates.
00:18
So let's first start off with this sketching using the inverse to be helpful.
00:23
What we need to remember is that the inverse of log base 2 of x, the inverse of this guy, is going to be an exponential.
00:32
So we know that if we were to look at y equals 2 to the x, it would be really easy to make the table as powers of 2 because we would have x and y and if we plug in negative 1 we get 2 to the negative first which is 1 half.
00:49
Then we could plug in 0, 2 to the 0 is 1.
00:52
Plug in 1 we get 2.
00:54
2 squared is 4.
00:56
2 cubed is 8.
00:58
2 to the 4th is 16.
01:01
2 to the 5th is 32 and so on.
01:05
One of the cool things here about using the graph of an exponential or a table for the exponential to get the graph of a log is that because they are inverses we are just swapping the x and the y values.
01:21
So if i swap the x and the y values this becomes log base 2 of x and we would have the points 1 half, negative 1, 1, 0, 2, 1, 4, 2, 8, 3, 16, 4, and 32, 5.
01:45
So let's plot these points.
01:47
We have 1 half, negative 1, 1, 0, 2, 1, 4, 2, 8, 3, and then 16, 4 and see that because the log function grows forever but grows slowly we are not going to be able to really get much height here on our graph unless we were to zoom way out on our x axis.
02:16
So here is what log base 2 of x looks like and we used our 2 to the x table to then invert to get our inverse so that we have our log base 2 to the x table which we then graphed right here.
02:31
So then we want to approximate the solutions to these equations down here.
02:39
So this first one is pretty straight forward.
02:41
If we divide both sides by 6 we get log base 2 of x equals 14 over 6 which is about 2 and a third.
02:52
So not about 2 and a third, it is 2 and a third.
02:55
So this comes out to be 2 .3.
03:00
So if we come up here and look at 2 .3 as a height value right here it looks like this is going to intersect at about 4 and a half.
03:14
So our x value here is about 4 .5 as our solution.
03:21
Moving on to part b.
03:23
If we rewrite this so that we have divided both sides by 2 we get log base 2 of 6x minus 21 equals 7 over 2 which is 3 and a half.
03:35
So the same kind of idea here.
03:37
We are going to go up to 3 and a half and put that in here on our height value and it looks like we intersect right here at 12.
03:48
So it looks like what goes inside here needs to be 12.
03:52
So we do 6x minus 21 equals 12.
03:56
If we add 21 we get 6x equals 33 and then divide by 3 so x would be 33 over 6 or approximately 33 over 6 because we are not quite sure that that 12 is exactly right.
04:18
And you know we could get a value for that 33 over 6 comes out to be a decimal value that you can report if you really want that.
04:31
A 5 and a half so you know it's about 5 and a half.
04:35
And then let's move on to part c.
04:37
Part c though and let's switch to a different dotted color.
04:41
Let's make this green here.
04:45
If we divide both sides by 7 here 15 divided by 7 comes out to give us 2 .14.
04:57
So that's going to be a little bit less than this guy right here 2 .14 and it's not going to let me jump to that.
05:05
So i'm going to draw it down here and then we can move it.
05:10
So 2 .14 i guess it really just doesn't want it to be there.
05:17
I guess we'll hand do it.
05:24
2 .14 would be under the red line so it's going to be a little bit less than a little bit more than 4 so maybe about 4 .25.
05:35
So we know that this number in here ends up being about 4 .25.
05:41
So now we have x squared minus 2x plus 1 equals 4 .25.
05:47
Let's move that 4 .25 over so x squared minus 2x minus 3 .25 equals 0.
05:54
And we're going to need the quadratic formula here to help us solve this.
05:57
So using the quadratic formula we get 2 plus or minus the square root of 17 over 2.
06:07
And if we approximate that we get about 3 or negative 1.
06:14
So those are our approximations.
06:16
So again let's put them up here.
06:19
We did this guy was about 11 .5 and then this guy gave us 3 or negative 1.
06:29
So let's go through here and actually solve these.
06:32
We have 6 times log base 2 of x equals 14.
06:41
I'm going to rewrite all of these real quick.
06:43
2 log base 2 of 6x minus 21 equals 7.
06:48
And then 7 log base 2 of x squared minus 2x plus 1 equals 15.
07:01
And we will probably move this down when we need more room.
07:06
And let's see here.
07:07
Let's go ahead and jump into solving this algebraically.
07:11
So the first thing that we'll do is we'll divide both sides by 6...