# 3 ques. Cumulative (logs,circles w/tangent lines)

1. Aug 3, 2009

### whitehorsey

1. Simplify each expression:
2 log49 - log23

2. none

3. 2 log49 - log23
= log481 - log23
= - log23
I get - log23 when it should just be log23. What did i do wrong?

1. Find the equation of the following circle:
The circle that passes through the origin and has intercepts equal to 1 and 2 on the x- and y- axes, respectively.

2. none

3. I have no idea how to do this problem.

1. For the circle x2 + y2 + 6x - 4y + 3 =0, find:
The equation of the tangent line at the point (-2,5).

2. xx0+yy0 = r2

3. I know that the center is (-3,2) and the radius is radical 10, but when i solve this problem I get the wrong answer.
xx0+yy0 = r2
x(-2) + y(5) = r2
-2x +5y = 10
The answer should be x + 3y = 13.
What did I do wrong?

Thank You!!

2. Aug 3, 2009

### songoku

Hi whitehorsey
Can you elaborate how you got - log23 ?

The equation of circle : (x - a)^2 + (y - b)^2 = r^2, where (a, b) is the center.
The circle passes through (0,0) ; (1,0) and (0,2)

That's the formula if the center is origin.

Find the slope first. Do you know implicit differentiation?

3. Aug 4, 2009

### HallsofIvy

If x= $log_4 81$ then $81= 4^x= (2^2)^x= 2^{2x}$ so $log_2 81= 2x$ and $x= log_4 81= (1/2)log_2 81= log_2 9$.
What is $log_2 9- log_2 3$?

$(x- a)^2+ (y- b)^2= R^2$ must be true for x= 0, y= 0; x= 1, y= 0; and x= 0, y= 2. You have $(0- a)^2+ (0- b)^2= a^2+ b^2= R^2$, $(1- a)^2+ b^2= R^2$, and $(0-a)^2+ (2- b)^2= R^2$. Three equations to solve for the three unknown values, a, b, and R. What happens if you subtract the first equation from the second? The first equation from the third?

You certainly don't need Calculus, as songoku suggested. You know the center of the circle is (-3, 2). What is the slope of the line from (-3, 2) to (-2, 5), a radius of the circle? The tangent at that point on the circle must be perpendicular to the radius: if two lines of slopes m1 and m2 are perpendicular, then m1m2= -1 so you can find the slope of the tangent line. What is the equation of the line through (-2, 5) with that slope?

Last edited by a moderator: Sep 5, 2009
4. Aug 4, 2009

### songoku

Nice suggestion Mr. Hall ^^

5. Aug 5, 2009

### whitehorsey

For the first problem I don't get how you could put the x down there ($81= 4^x= (2^2)^x= 2^2x$) when it was an exponent before.

Thank You for your help on the second problem. I understand it now. ^ ^

For the third problem, I find the slope and it was 3. Then I tried solving for the tangent line but i ended up with y = 3x +11. Can you teach me what I did wrong? I tried different ways but I can't seem to figure it out.

6. Aug 7, 2009

### whitehorsey

i found the slope which is 3. What do i do next? hmmm i don't recall ever learning implicit differentation

7. Aug 7, 2009

### rock.freak667

x=log4 81 (you know that logaa=1 right?)

think of it like this xlog44=log481 =>log44x=log481 => 4x=81

Now doing what HallsofIvy said: 4=22 so we have (22)x=22x.
you can continue from where HallsofIvy left of.

Another way is to use the 'change of base formula' for logs.

$$log_a b = \frac{log_c b}{log_c a}$$

So for 2 log49 - log23 , change either of the logs base to the other and simplify (base 4 to 2 or vice versa)

The angle between a radius and a tangent at the point of contact is 90 degrees i.e. radius and tangent are perpendicular.

Now if line 1 and line 2 (with gradients m1 and m2 respectively) are perpendicular then m1m2=-1 (as HallsofIvy said previously)

the gradient you found is the gradient of the radius connecting the centre to (-2,5). Can you then find the gradient of the tangent? When you get that, you have the gradient of the tangent and a point on the tangent, you can now find the required equation.

8. Aug 7, 2009

### whitehorsey

Hi rock.freak667
I was wondering how the change of base formula allows you to change the 4 into the 2 or vice versa like you said. Because I thought it only allows you to solve problems like these log37 which would change into log 7/log 3.

For the other question, I have tried many different ways but cannot seem to figure out the gradient of the tangent. Can you please show me? I'm very confused right now. :[

9. Aug 7, 2009

### mathie.girl

When you changed log37 into log 7/log 3 you actually started using a new base of 10 (remember if it doesn't have a base written there, you're actually working in base 10). This is often done because your calculator has a button with this base on it and it allows you to use your calculator to compute this log. However, you could have chosen any base. For instance,

$$log_{3}7 = \frac{log_{a}7}{log_{a}3} = \frac{log_{15}7}{log_{15}3} = \frac{log_{957}7}{log_{957}3}$$

Normally you'd pick $$a = 10$$, but if you want to work with base 2, you can also choose that.

For the slope of the tangent, you need to use that the radius of a circle and the tangent to a circle at the same point are perpendicular. Do you remember about the slopes of perpendicular lines? If line $$l$$1 is perpendicular to $$l$$2, their slopes are negative reciprocals. So pretty much, you take the slope $$m$$ of $$l$$1, make it negative and flip it over in order to get the slope of $$l$$2.

10. Sep 4, 2009

### whitehorsey

I got it Thank You!!

11. Sep 5, 2009

### HallsofIvy

I didn't. That was a typo on my part. I mean $81= 4^x= (2^2)^x= 2^{2x}$

The slope of a radius, from (-3,2) to (-2,5), is 3. As I said, a tangent must be perpendicular to that radius and so has slope -1/3.