What are the repeating digits in the smallest group of repeating digits in the decimal equivalent of 28/11?
2 answers:
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Answer:
h(x) = 4·log₃(x) +2
Step-by-step explanation:
<h3>Part A:</h3>h(x) = f(x) +g(x)
h(x) = (log₃(x) +3) +(log₃(x³) -1)
h(x) = log₃(x) +3·log₃(x) +2
h(x) = 4·log₃(x) +2 . . . . . "simplest" form
h(x) = log₃(9x⁴) . . . . . . . as a single logarithm
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<h3>Part B:</h3>No system of equations is given. Perhaps you want to find x for f(x) = g(x).
log₃(x) +3 = log₃(x³) -1
log₃(x) +3 = 3·log₃(x) -1
4 = 2·log₃(x) . . . . . . . . . . . add 1-log₃(x)
2 = log₃(x) . . . . . . . . . . . . divide by 2
3² = x = 9 . . . . . . . . . . . . . take the antilog
The solution to f(x) = g(x) is x = 9.
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<em>Additional comment</em>
The relevant rules of logarithms are ...
a = log₃(b) ⇔ 3^a = b
log(a^b) = b·log(a)
log(ab) = log(a) +log(b)
