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:
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
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