answer:We have 27sqrt3 = (3^3)(3^frac12)=3^{(3+frac12)}=3^{frac72}. Therefore, log_3 27sqrt3=frac72.

answer:I need to find the inverse function of f, which means I need to switch the roles of x and y and then solve for y. So, if f(x)=2x^3+4, then y=2x^3+4, and swapping x and y gives x=2y^3+4. Now I need to isolate y on one side of the equation. First, I'll subtract 4 from both sides to get x-4=2y^3. Next, I'll divide both sides by 2 to get frac{x-4}{2}=y^3. Finally, I'll take the cube root of both sides to get y=sqrt[3]{frac{x-4}{2}}. This is the inverse function of f. To find f^{-1}(58), I just need to plug in x=58 into the inverse function and simplify. So, f^{-1}(58)=sqrt[3]{frac{58-4}{2}}=sqrt[3]{27}=3.

answer:To solve this problem, I need to find the common difference of the arithmetic sequence, which is the amount by which each term increases from the previous one. I also need to use the fact that the product of two consecutive terms is 13, which is a prime number. This means that one of the terms must be 1 and the other must be 13, since those are the only factors of 13. Since the sequence is increasing, I can assume that a_4 = 1 and a_5 = 13. Now, to find the common difference, I can subtract any two consecutive terms. For example, a_5 - a_4 = 13 - 1 = 12. So the common difference is 12, and I can use this to find a_3 and a_6. To find a_3, I can subtract 12 from a_4, which gives me a_3 = a_4 - 12 = 1 - 12 = -11. To find a_6, I can add 12 to a_5, which gives me a_6 = a_5 + 12 = 13 + 12 = 25. Now, I have a_3 = -11 and a_6 = 25, so I can multiply them to get a_3a_6 = -11 times 25 = -275.

answer:I notice that both equations have the factor (x+y), so I try to eliminate it by adding or subtracting the equations. If I add them, I get (x+y)(x+y) = 27 + 54, which simplifies to (x+y)^2 = 81. This is exactly what I need to find, so I don't need to do anything else.