how does one solve: 2+(3+5y^4)^6

Question

the order of operations says to solve what is within the parenthesis first but the rules of exponents says we would handle the outside exponent (6) first. 

Answer 1

The Order of Operations is:

  Parentheses

  Exponentiation

  Multiplication, Division    -  left to right

  Addition, Subtraction     -  left to right

 

This mean that the expression is evaluated as follows:

  2 + ( 3 + 5y⁴ )⁶       [evaluate expression in Parentheses first -->   3+5y⁴ ]

                       3 + 5y⁴      [Recursively use P-E-M-D-A-S; no more P, so do Exponentiation]

                       3 + 5a       [I'll use 'a' since we don't have value for y; do Multiplication]

                       3 + b        [using b]

                          c            [do Addition; result c]

  2 + c⁶    [no more Parentheses; now do Exponentiation]

     d        [d is the result of Addition]

 

Now, PLZ note that the "Rules of Exponents" says that y⁴ =y*y*y*y and c⁶ = c*c*c*c*c*c.  Where else would use use these rules??

Answer 2

"Solve" is a direction to determine the value of the variable that makes the equation true.

You present an expression, not an equation. It cannot be solved.

The expression can be expanded.

The exponent 6 must be done first.

2 + (3 + 5y⁴)(3 + 5y⁴)(3 + 5y⁴)(3 + 5y⁴)(3 + 5y⁴)(3 + 5y⁴)