Question

I need help with pre calculus.

Answer 1

h(x) = 4·log₃(x) +2

Explanation:

Part A:

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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Part B:

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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Additional comment

The relevant rules of logarithms are ...

a = log₃(b) ⇔ 3^a = b

log(a^b) = b·log(a)

log(ab) = log(a) +log(b)