A: It is easier (though not always necessary) to have all of the logs in the same base to proceed. Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:
log_{a}b = log_{c}b / log_{c}a where c can be any base of your choosing

What base should we go to? Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

Q: Solve the system of equations (both of which are conic sections)
1: x^{2} + y^{2} – 20x + 8y + 7 = 0
2: 9x^{2} + y^{2} + 4x + 8y + 7 = 0

A: I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2: 9x^{2} + y^{2} + 4x + 8y + 7 = 0

1: -x^{2} – y^{2} + 20x – 8y – 7 = 0 [after it has been multiplied by -1]

Now, add them together to get equation 3:

3: 8x^{2} +24x = 0

Now, equation 3 only has x’s so we can solve for x (by factoring):

(8x)(x + 3) = 0

So, x = 0 or -3

But, we aren’t done… Find the solution to this system of conics means that we are finding the point(s) of intersection — and it turns out there are two points of intersection since there were two solutions for x. So, our points (solutions) are:

(0, ?) and (-3, ?)

We need to find the y-coordinate of each solution separately. We do this by plugging in the x value to either of the original equations (both will lead us to the same correct solution):

First x-value: (0, ?)

x^{2} + y^{2} – 20x + 8y + 7 = 0

(0)^{2} + y^{2} – 20(0) + 8y + 7 = 0

y^{2} + 8y + 7 = 0

Solve for y by factoring:

(y + 7)(y + 1)=0

y = -7, -1

OH! So this really means that there are more than two solutions…. When x = 0, we found two answers for y.

(0, -7) and (0, -1)

Second x-value: (-3, ?)

x^{2} + y^{2} – 20x + 8y + 7 = 0

(-3)^{2} + y^{2} – 20(-3) + 8y + 7 = 0

9 + y^{2} +60 + 8y + 7 = 0

y^{2} + 8y + 76 = 0

Since this does not factor, we need to solve for y by using the quadratic formula. I did not show my work here, but we end up with a negative under the square root, so there are no real solutions for when x = -3.

Therefore, there are two real solutions to this equation (2 points of intersection) and they are:

(0, -7) and (0, -1)

Here is a visual of what we are solving for….. We had the equations for the two graphed conic sections and we found the two points of intersection shown below:

A: There are a few ways to do this, but first I will rationalize the denominator (that means get rid of the square root in the denominator) and then I will simplify the square root (simplest radical form):

Step 1: Rationalize the denominator:

Multiply the top and bottom by sqrt(75):

Now, simplify algebraically:

Simplify the denominator to get 150… for some reason, I can’t make my picture keep it in a fully fraction with 150 in the bottom, so the image shows the 150 pulled out like so (either way is fine):

Now reduce the 15 and the 150 to get:

The denominator has been rationalized (no square roots in the denominator) and the fraction has been reduced.

Step 2: Put the numerator into simplest radical form:

To go to simplest radical form, you want to see what “perfect squares” divide into the square root. So, what perfect squares divide into 75? 25 is a perfect square that divides into 75: 3*25=75.

So, we can break up the square root like so (again, the square roots can be on top of the fraction, just having image problems):

Now, what is the square root of 25? That’s 5, so simplify:

And, now reduce the fraction part of the 5 and the 10 in the denominator to get: