Question

Log₂((3,(7)+3^(-2) +0!÷9)×√((3^((5!!-6!!)×(-1)))÷3^27)+40×sin30°)=

Answer 1

0!=1
0!:9=1/9
3,(7)=3 +(0,7+0,07+0,007+...)=3 целых 7/9

log₂((3,(7)+3⁻² +0!÷9)=log₂(3 целых 7/9+(1/9)+(1/9))=log₂4=2

5!!-6!!=1*3*5-(2*4*6)=15-48=-33

3³³: 3²⁷ =3⁶

40×sin30°=40·(1/2)=20


log₂((3,(7)+3^(-2) +0!÷9)×√((3^((5!!-6!!)×(-1)))÷3^27)+40×sin30°)=

=2*√(3⁶+20)=2√749

Answer 2

Let's break down the expression step by step:

1. Start by solving the innermost operations in parentheses and exponentiation: a. (3,7) = 3^7 (comma operator is a shorthand for exponentiation) b. 3^(-2) = 1/(3^2) c. 0! = 1 (by definition, the factorial of 0 is 1) d. 1 ÷ 9 = 1/9

   The expression now becomes: Log₂((3^7 + 1/(3^2) + 1/9) × ...

2. Next, let's deal with the factorial and double factorial: a. 5!! = 5 × 3 × 1 = 15 b. 6!! = 6 × 4 × 2 = 48 c. 5!! - 6!! = 15 - 48 = -33

   The expression now becomes: Log₂((3^7 + 1/(3^2) + 1/9) × √((3^(-33×(-1))) ÷ 3^27) + 40 × sin(30°))

3. Continue with exponentiation: a. 3^(-33×(-1)) = 3^33

   The expression now becomes: Log₂((3^7 + 1/(3^2) + 1/9) × √((3^33) ÷ 3^27) + 40 × sin(30°))

4. Simplify the square root: a. √(3^33 ÷ 3^27) = √(3^(33-27)) = √(3^6) = 3^3 = 27

   The expression now becomes: Log₂((3^7 + 1/(3^2) + 1/9) × 27 + 40 × sin(30°))

5. Calculate the remaining parts: a. 3^7 = 2187 b. 1/(3^2) = 1/9 c. 1/9 = 0.111111111... d. 40 × sin(30°) = 40 × 0.5 = 20

   The expression now becomes: Log₂((2187 + 0.111111111... + 0.111111111...) × 27 + 20)

6. Simplify the constants: a. 0.111111111... (repeating decimal) is approximately equal to 1/9 ≈ 0.111111111... b. (2187 + 1/9 + 1/9) × 27 = (2187 + 2/9) × 27 ≈ 2187 × 27 ≈ 59049

   The expression now becomes: Log₂(59049 + 20)

7. Calculate the final result: a. 59049 + 20 = 59069 b. Log₂(59069) ≈ 15.84426

So, the final result of the expression is approximately Log₂(59069) ≈ 15.84426.