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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?

1 Answer

3 votes

Answer:

x = 8

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

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?-example-1
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