Question

Let f (x) = log₃(x) + 3 and g(x) = log₃(x³) – 1.

Part A: If h(x) = f (x) + g(x), solve for h(x) in simplest form.

Part B: Determine the solution to the system of nonlinear equations of f(x) and g(x).

Answer 1

See below

Explanation:

`\textbf{Part A: } \text{ If }h(x) = f (x) + g(x) \text{, solve for } h(x) \text{ in simplest form.}`

We have

`f (x) = \log_3(x) + 3`

and

`g(x) = \log_3(x^3) - 1`

Thus

`h(x) = f (x) + g(x) = (\log_3(x) + 3) + ( \log_3(x^3) - 1) = \log_3(x) + 3 + \log_3(x^3) - 1`

`= \log_3(x) + \log_3(x^3) + 2`

Recall the property of logarithms:

`\boxed{\log_b(n\cdot m) = \log_b(n) + \log_b(m)}`

then,

`\log_3(x) + \log_3(x^3) + 2 = \log_3(x\cdot x^3) +2 = \boxed{\log _3(x^4)+2}`

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`\textbf{Part B: } \text{Determine the solution to the system of nonlinear equations of } f(x) \text{ and } g(x).`

I am assuming that the system of equations is

`$\left \{ {{ f (x) = \log_3(x) + 3 } \atop {g(x) = \log_3(x^3) - 1}} \right$`

and you probably want the solution when
`f(x)=g(x)` I will name it
`y`, thus

`$\left \{ {{ f (x) = \log_3(x) + 3 } \atop {g(x) = \log_3(x^3) - 1}} \right \implies \left \{ {{y= \log_3(x) + 3 } \atop {y = \log_3(x^3) - 1}} \right$`

We should just solve

`\log_3(x) + 3 = \log_3(x^3) - 1`

`\log_3(x) - \log_3(x^3) = - 4`

`\log_3 \left((x)/(x^3) \right) = - 4`

`\log_3 \left((1)/(x^2) \right) = - 4`

`\log_3 (x^(-2)) = - 4`

`-2\log_3 (x) = - 4`

`\log_3 (x) = 2 \iff 3^2 = x \implies \boxed{x= 9}`