Respuesta :

InesWalston

Answer-

The domain of [tex]g(x)=3\log 2(x-1)+4[/tex] is,

[tex]\boxed{\boxed{(x:x\ \epsilon\ R\ and\ R>1)}}[/tex]

Solution-

For [tex]y=\log x[/tex], there can not have a log to any value of x, which is either negative or 0.

So domain of log x is,

[tex](x:x\ \epsilon\ R\ and\ R>0)[/tex]

Equating 2(x-1) to 0, we can get the domain of the given function, hence

[tex]\Rightarrow 2(x-1)>0\\\\\Rightarrow x-1>0\\\\\Rightarrow x>1[/tex]

Therefore, domain of g(x) is any real number greater than 1, i.e

[tex](x:x\ \epsilon\ R\ and\ R>1)[/tex]


LucianoBordoli

Answer:

All real numbers greater than 1

Step-by-step explanation:

The domain of a function is the set of possible values in which we can evaluate the function. In other words, it is the possible set of values for x in the expression of the function.

The function is :

[tex]g(x)=3log2(x-1)+4[/tex]

The only restriction in the domain of the function g(x) is given by the logarithm in the function.

The argument of the logarithm function must be greater than 0.

Given a logarithm function [tex]log_{a}b=c[/tex] the expression b is called the argument ⇒ In this exercise we need

[tex]2(x-1)>0[/tex]

[tex](x-1)>0[/tex]

[tex]x>1[/tex]

There isn't another restriction for the logarithm inside g(x) ⇒

The domain of g(x) is all real numbers greater than 1.

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