The3Mathketeers

I have to confess that I committed a serious error in my calculations because I forgot to consider that a valid even-numbered exponent is 0. This means that there is one additional choice for each prime factor 2, 3, and 5. \(5 * 3 * 3 = 45 \) perfect square factors! I apologize in advance for this mistake!

I apologize for the confusion. As you discovered, the diagram I created is intentionally NOT drawn to scale so that it looks as similar as possible to the original diagram. I figured that as long as points E and F were on the circle, then I could still go ahead and find the radius anyway.

First, perform the factorization procedure on 60. I am not sure how to show this properly, so I will do my best. I use the red color to distinguish a prime number factor.

\(60 = \textcolor{red} 2 * 30 \\ 30 = \textcolor{red} 2 * 15 \\ 15 = \textcolor{red} 3 * \textcolor{red} 5\)

Now, that we have found the prime factorization, we know that \(60 = 2^2 \times 3 \times 5\). However, we need the factorization of \(60^4\). We can accomplish this by doing the following:

\(60 = 2^2 \times 3 \times 5 \\ 60^4 = \left(2^2 \times 3 \times 5\right)^4 \\ 60^4 = 2^8 \times 3^4 \times 5^4\)

Now, we have the prime factorization of 60^4. In order for a number to be a perfect square factor, the factor would need an even-numbered power of all the factors 2, 3, and 5.

For the factor 2, the even-numbered powers possible are 2, 4, 6, and 8 for a total of 4 choices.

For the factor 3, the even-numbered powers possible are 2 and 4 for a total of 2 choices.

For the factor 5, the even-numbered powers possible are 2 and 4 for a total of 2 choices.

These choices are completely independent choices, so we can use the multiplication rule to find the number of perfect square factors of \(60^4 \). The number of perfect square factors is \(4 * 2 * 2 = 16\) perfect square factors.

I had to stare at this one a long time to solve this one! I have created a diagram below with points labeled so that it is easier to follow along.

In the diagram above, notice that OF and OE are both radii of the semicircle. These lengths I have labeled as r. I have labeled L as the length of segment OB. We can now apply Pythagorean's Theorem twice to find the radius of this semicircle.

Now, we have expressions for both r and L, so we can solve for r, and we are finished. Probably the best way is to solve for L first since that one is easier to solve.

\(2^2 + (2 + l)^2 = 5^2 + (5 - l)^2 \\ 4 + 4 + 4l + l^2 = 25 + 25 - 10l + l^2 \\ 8 + 4l = 50 - 10l \\ 14l = 42 \\ l = 3\)

Now that we have solved for L, we can find the radius.

\(r^2 = 2^2 + (2 + l)^2 \\ r^2 = 2^2 + (2 + 3)^2 \\ r^2 = 4 + 25 \\ r^2 = 29 \\ r = \sqrt{29} \text{ or } r = -\sqrt{29}\)

Since we are dealing with the radius, which is a length, we should reject \(r = -\sqrt{29}\). Therefore, \(r = \sqrt{29} \approx 5.3852\)

I did some research after posting this answer, and I am relatively certain that the answer I posted previously is not correct, but I will leave this here in case it helps someone else solve this problem. In a polynomial in the form \(P(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_2x^2 + a_1x + a_0\), it seems like there is some ambiguity about whether or not \(a_0\) fits the definition of a coefficient or not. I always considered \(a_0\) as classified as a constant, but some argue that \(a_0\) is a coefficient of the \(x^0\) term. If that interpretation fits this problem, then the answer I posted previously is not correct. I made a few attempts, but I was unable to solve this problem with the new interpretation.

To rationalize an expression such as \(\frac{1}{2 - \sqrt[3]{2}}\), you first recognize that there is a cube root in the denominator. This immediately provides insight as to how to approach this particular problem. Personally, I find working with fractional exponents is far easier than radicals, so I will almost always convert to this.

Since there is a cube root within the denominator, we will have to make the denominator a perfect cube in order to ensure that the denominator has no radicals. We can do this by making a key observation: \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\). This is the formula for the difference of a perfect cube. This information should be enough to rationalize this problem and ones like it. 

If we look closely at the denominator, we can match up the formula for a perfect cube to the denominator of this particular problem.

If we let \(A = 2\) and \(B = \sqrt[3]{2} = 2^\frac{1}{3}\), then that would make the following identity. 

\(A = 2; B = 2^\frac{1}{3} \\ 2^3 - \left(2^\frac{1}{3}\right)^3 = \left(2 - 2^\frac{1}{3}\right)\left(2^2 + 2*2^\frac{1}{3} + \left(2^\frac{1}{3}\right)^2\right) \\ 6 = \left(2 - 2^\frac{1}{3} \right) \left(4 + 2^\frac{4}{3} + 2^\frac{2}{3}\right) \\\)

Now, we can rationalize because we have the term that will cancel out all radicals within the denominator.

\(\frac{1}{2 - 2^\frac{1}{3}} * \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{4 + 2^\frac{4}{3} + 2 ^\frac{2}{3}} = \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6}\)

Now, we have to convert to the radical notation and figure out how it matches the desired format of \(\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}\).

\(\frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6} = \frac{4 + \sqrt[3]{2^4} + \sqrt[3]{2^2}}{6} = \frac{\sqrt[3]{64} + \sqrt[3]{16} + \sqrt[3]{4}}{6}\)

Now, we identify the desired variables and find their sum.

\(A = 64, B=16, C=4, D=6 \\ A + B + C + D = 64 + 16 + 4 + 6 = 90\)

I think this problem is easier than it may first appear unless I am misreading the question.

First, I would consider the given information and determine if I can fit all the criteria in a first-degree polynomial. I think I can. Just do this: \(P(x) = x - \left(\sqrt[3]{2} + \sqrt[3]{3}\right)\). This is a first-degree polynomial, so this must be of least degree, and P(x) has rational coefficients. The coefficients of the x-term is 1, which is rational. Because of the way I have written this polynomial, it is guaranteed to have the root of \(\sqrt[3]{2} + \sqrt[3]{3}\).

There is only one root, so the product of the roots is just \(\sqrt[3]{2} + \sqrt[3]{3}\).

Simply use some creative algebraic manipulation with the property of logarithms to find an approximation for the other expressions.

a) \(\log_{5}2\)

There are probably many approaches that would lead to the same answer, but here is how I would approach this one:

\(\log_{5}8 =\log_{5}2^3 \\ \log_58 = 3\log_52 \\ 1.2920 = 3\log_52 \\ \log_52 = \frac{1.2920}{3} \approx 0.4307\)

b) \(\log_5 \frac{75}{8}\)

Luckily, we use the result from before to help this particular problem.

\(\log_5 \frac{75}{8} = \log_5 75-\log_5 8 \\ \log_5 \frac{75}{8} = \log_5 (3*25) - \log_5 8\\ \log_5 \frac{75}{8} = \log_5 3 + \log_5 25 - \log_5 8\\ \log_5 \frac{75}{8} = 0.6826 + 2 - 1.2920 \approx 1.3906 \\ \)

You can check to determine if your answer is reasonable by using a calculator and evaluating \(\log_5 2\) and \(\log_5 \frac{75}{8}\). The decimal expansion should be match up for the first few digits.