Math / Algebra

QuestionWrite the domain in interval notation. $k(x)=\ln \left(x^{2}+7\right)$
The domain is $\square$ .

Studdy Solution

STEP 1

What is this asking? We need to find all the x values that we can plug into the function $k(x) = \ln(x^2 + 7)$ without breaking any math rules! Watch out! Remember that we can only take the natural log ( $\ln$ ) of positive numbers.
Zero is not allowed!

STEP 2

1. Inside the Logarithm
2. Solve the Inequality
3. Interval Notation

STEP 3

Alright, so we've got this natural log function, and we know the inside of the natural log must be greater than zero.
So, let's set up our inequality: $x^2 + 7 > 0$

STEP 4

Let's isolate $x^2$ by subtracting 7 from both sides: $x^2 > -7$

STEP 5

Now, think about this: any real number squared will always be greater than or equal to zero.
Since zero is already bigger than -7, any number we plug in for x will work! $x^2$ will always be positive or zero, and therefore always greater than -7.

STEP 6

Since any real number works for x, our domain is all real numbers.
In interval notation, that's written as: $(-\infty, \infty)$ Boom!

STEP 7

The domain of $k(x)$ is $(-\infty, \infty)$ .

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QuestionWrite the domain in interval notation. $k(x)=\ln \left(x^{2}+7\right)$
The domain is $\square$ .

Studdy Solution

STEP 1

What is this asking? We need to find all the x values that we can plug into the function $k(x) = \ln(x^2 + 7)$ without breaking any math rules! Watch out! Remember that we can only take the natural log ( $\ln$ ) of positive numbers.
Zero is not allowed!

STEP 2

1. Inside the Logarithm
2. Solve the Inequality
3. Interval Notation

STEP 3

Alright, so we've got this natural log function, and we know the inside of the natural log must be greater than zero.
So, let's set up our inequality: $x^2 + 7 > 0$

STEP 4

Let's isolate $x^2$ by subtracting 7 from both sides: $x^2 > -7$

STEP 5

Now, think about this: any real number squared will always be greater than or equal to zero.
Since zero is already bigger than -7, any number we plug in for x will work! $x^2$ will always be positive or zero, and therefore always greater than -7.

STEP 6

Since any real number works for x, our domain is all real numbers.
In interval notation, that's written as: $(-\infty, \infty)$ Boom!

STEP 7

The domain of $k(x)$ is $(-\infty, \infty)$ .

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QuestionWrite the domain in interval notation. k(x)=ln⁡(x2+7)k(x)=\ln \left(x^{2}+7\right)k(x)=ln(x2+7)The domain is □\square□ .

Studdy Solution

STEP 1

What is this asking? We need to find all the *x* values that we can plug into the function k(x)=ln⁡(x2+7)k(x) = \ln(x^2 + 7)k(x)=ln(x2+7) without breaking any math rules! Watch out! Remember that we can only take the natural log (ln⁡\lnln) of *positive* numbers.Zero is not allowed!

STEP 2

1. Inside the Logarithm2. Solve the Inequality3. Interval Notation

STEP 3

Alright, so we've got this natural log function, and we know the *inside* of the natural log *must* be greater than zero.So, let's **set up** our inequality: x2+7>0x^2 + 7 > 0x2+7>0

STEP 4

Let's **isolate** x2x^2x2 by subtracting 7 from both sides: x2>−7x^2 > -7x2>−7

STEP 5

Now, think about this: *any* real number squared will *always* be greater than or equal to zero.Since zero is already bigger than -7, *any* number we plug in for *x* will work! x2x^2x2 will always be positive or zero, and therefore always greater than -7.

STEP 6

Since *any* real number works for *x*, our domain is all real numbers.In interval notation, that's written as: (−∞,∞)(-\infty, \infty)(−∞,∞) Boom!

STEP 7

The domain of k(x)k(x)k(x) is (−∞,∞)(-\infty, \infty)(−∞,∞).

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QuestionWrite the domain in interval notation. $k(x)=\ln \left(x^{2}+7\right)$
The domain is $\square$ .

What is this asking? We need to find all the x values that we can plug into the function $k(x) = \ln(x^2 + 7)$ without breaking any math rules! Watch out! Remember that we can only take the natural log ( $\ln$ ) of positive numbers.
Zero is not allowed!

1. Inside the Logarithm
2. Solve the Inequality
3. Interval Notation

Alright, so we've got this natural log function, and we know the inside of the natural log must be greater than zero.
So, let's set up our inequality: $x^2 + 7 > 0$

Let's isolate $x^2$ by subtracting 7 from both sides: $x^2 > -7$

Now, think about this: any real number squared will always be greater than or equal to zero.
Since zero is already bigger than -7, any number we plug in for x will work! $x^2$ will always be positive or zero, and therefore always greater than -7.

Since any real number works for x, our domain is all real numbers.
In interval notation, that's written as: $(-\infty, \infty)$ Boom!

The domain of $k(x)$ is $(-\infty, \infty)$ .