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

Answer:

See below

Step-by-step 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

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

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

$\log_3 \left(\dfrac{x}{x^3} \right) = - 4$

$\log_3 \left(\dfrac{1}{x^2} \right) = - 4$

$\log_3 (x^{-2}) = - 4$

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

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