Math / Algebra

Question $=1$ $=2$ $=3$ 4 5 6 7 8 9
Write the domain in Interval notation. $g(x)=\log (3-x)$
The domain is $\square$ .

Studdy Solution

STEP 1

1. The function $g(x) = \log(3-x)$ involves a logarithm.
2. The domain of a logarithmic function is the set of all $x$ such that the argument of the logarithm is positive.

STEP 2

1. Identify the argument of the logarithm.
2. Set up the inequality for the argument to be positive.
3. Solve the inequality.
4. Write the solution in interval notation.

STEP 3

Identify the argument of the logarithm in the function $g(x) = \log(3-x)$ .
The argument is $3-x$ .

STEP 4

Set up the inequality for the argument to be positive:
$3-x > 0$

STEP 5

Solve the inequality $3-x > 0$ :
$3 > x$
This can also be written as:
$x < 3$

STEP 6

Write the solution in interval notation. Since $x$ must be less than 3, the interval is:
$(-\infty, 3)$
The domain of the function $g(x) = \log(3-x)$ is $\boxed{(-\infty, 3)}$ .

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Question $=1$ $=2$ $=3$ 4 5 6 7 8 9
Write the domain in Interval notation. $g(x)=\log (3-x)$
The domain is $\square$ .

Studdy Solution

STEP 1

1. The function $g(x) = \log(3-x)$ involves a logarithm.
2. The domain of a logarithmic function is the set of all $x$ such that the argument of the logarithm is positive.

STEP 2

1. Identify the argument of the logarithm.
2. Set up the inequality for the argument to be positive.
3. Solve the inequality.
4. Write the solution in interval notation.

STEP 3

Identify the argument of the logarithm in the function $g(x) = \log(3-x)$ .
The argument is $3-x$ .

STEP 4

Set up the inequality for the argument to be positive:
$3-x > 0$

STEP 5

Solve the inequality $3-x > 0$ :
$3 > x$
This can also be written as:
$x < 3$

STEP 6

Write the solution in interval notation. Since $x$ must be less than 3, the interval is:
$(-\infty, 3)$
The domain of the function $g(x) = \log(3-x)$ is $\boxed{(-\infty, 3)}$ .

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Question=1=1=1 =2=2=2 =3=3=3 4 5 6 7 8 9Write the domain in Interval notation. g(x)=log⁡(3−x)g(x)=\log (3-x)g(x)=log(3−x)The domain is □\square□ .

Studdy Solution

STEP 1

1. The function g(x)=log⁡(3−x) g(x) = \log(3-x) g(x)=log(3−x) involves a logarithm.2. The domain of a logarithmic function is the set of all x x x such that the argument of the logarithm is positive.

STEP 2

1. Identify the argument of the logarithm.2. Set up the inequality for the argument to be positive.3. Solve the inequality.4. Write the solution in interval notation.

STEP 3

Identify the argument of the logarithm in the function g(x)=log⁡(3−x) g(x) = \log(3-x) g(x)=log(3−x).The argument is 3−x 3-x 3−x.

STEP 4

Set up the inequality for the argument to be positive:3−x>0 3-x > 0 3−x>0

STEP 5

Solve the inequality 3−x>0 3-x > 0 3−x>0:3>x 3 > x 3>xThis can also be written as:x<3 x < 3 x<3

STEP 6

Write the solution in interval notation. Since x x x must be less than 3, the interval is:(−∞,3) (-\infty, 3) (−∞,3) The domain of the function g(x)=log⁡(3−x) g(x) = \log(3-x) g(x)=log(3−x) is (−∞,3)\boxed{(-\infty, 3)}(−∞,3)​.

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Question $=1$ $=2$ $=3$ 4 5 6 7 8 9
Write the domain in Interval notation. $g(x)=\log (3-x)$
The domain is $\square$ .

1. The function $g(x) = \log(3-x)$ involves a logarithm.
2. The domain of a logarithmic function is the set of all $x$ such that the argument of the logarithm is positive.

1. Identify the argument of the logarithm.
2. Set up the inequality for the argument to be positive.
3. Solve the inequality.
4. Write the solution in interval notation.

Identify the argument of the logarithm in the function $g(x) = \log(3-x)$ .
The argument is $3-x$ .

Set up the inequality for the argument to be positive:
$3-x > 0$

Solve the inequality $3-x > 0$ :
$3 > x$
This can also be written as:
$x < 3$

Write the solution in interval notation. Since $x$ must be less than 3, the interval is:
$(-\infty, 3)$
The domain of the function $g(x) = \log(3-x)$ is $\boxed{(-\infty, 3)}$ .