Submit your math problems to me at mathoutloud40@gmail.com and I'll attempt a solution as I see it for the first time. Follow me on Twitter: / mathoutloud40 Follow me on Mastodon: mathstodon.xyz... #math #maths #mathematics
For (D) can we not just say 2^5 = 32 so log2 (30) < 5 and 3^4 = 81 so log3(85) > 4 so smaller/bigger means overall it's definitely less than 5/4. For (C) you forgot the squared on the 6!, but with all the cancelling what you get is comparing 7/4 to 7/(very big)
yea in the one which has square root of 10!, you forgot to write the squared, which was already given in the denominator, so squaring it again would make it ^4
I always have a lot of fun with problems like this. Here was my approach: I started by comparing the first two (since 5/4 is nice and easy)... sqrt(7)/2 is sqrt(28)/4. sqrt(28) is a little bit more than 5, so (A) must be more than (B). Looking at (C), I canceled a bunch of terms and got sqrt(7)/2160, which is a VERY small number, so obviously that's not it. For (D), I observed that log2(30) is slightly less than 5, and log3(85) is slightly more than 4, which means it is less than 5/4. (E) was the trickiest, but I wanted to compare it to (A)... so I set up the question sqrt(7)/2 (1 + sqrt(6))/3. I cross multiplied to get 3 sqrt(7) 2 + 2 sqrt(6). Then I moved the coefficients inside the square roots to get sqrt(63) 2 + sqrt(24). The left side is slightly less than 8 (but greater than 7), the right side is 2 + slightly less than 5. So that means the left side is greater. So the answer is (A).
