Question

If F(a, b, c, d) = a^b + c \times d, what is the value of x such that F(2, x, 4, 11) = 300?

Answer 1

Answer:

  x = 8

Step-by-step explanation:

A graphing calculator can show you the answer easily. It works well to define a function whose x-intercept is the solution. We can do that by subtracting 300 from the given equation so we have ...

  F(2, x, 4, 11) -300 = 0

The solution is x = 8.

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We can solve this algebraically:

  F(2, x, 4, 11) -300 = 0

  2^x +4·11 -300 = 0 . . . . use the function definition

  2^x -256 = 0 . . . . . . simplify

  2^x = 2^8 . . . . . add 256

  x = 8 . . . . . . . . . match exponents of the same base