Multiplicative Inverses Mod n Write a program which takes in two integers a, and n. Your job is to ensure that they are both positive integers (just to reinforce positivity, zero is not legal input either). Then your objective is to find out if there is an integer x so that a times x is 1 mod n. Sometimes they exist, and sometimes they won't. Test all possibilities between 1 and n to see if they work.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places