Trigonometry Examples

f (x) = ln (2) + ln (x - 3)

$f (x) = \ln (2) + \ln (x - 3)$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in

0

$0$ for

y

$y$ and solve for

x

$x$ .

0 = ln (2) + ln (x - 3)

$0 = \ln (2) + \ln (x - 3)$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as

ln (2) + ln (x - 3) = 0

$\ln (2) + \ln (x - 3) = 0$ .

ln (2) + ln (x - 3) = 0

$\ln (2) + \ln (x - 3) = 0$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Use the product property of logarithms,

{log}_{b} (x) + {log}_{b} (y) = {log}_{b} (x y)

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

ln (2 (x - 3)) = 0

$\ln (2 (x - 3)) = 0$

Step 1.2.2.2

Apply the distributive property.

ln (2 x + 2 \cdot - 3) = 0

$\ln (2 x + 2 \cdot - 3) = 0$

Step 1.2.2.3

Multiply

2

$2$ by

- 3

$- 3$ .

ln (2 x - 6) = 0

$\ln (2 x - 6) = 0$

ln (2 x - 6) = 0

$\ln (2 x - 6) = 0$

Step 1.2.3

To solve for

x

$x$ , rewrite the equation using properties of logarithms.

e^{ln (2 x - 6)} = e^{0}

$e^{\ln (2 x - 6)} = e^{0}$

Step 1.2.4

Rewrite

ln (2 x - 6) = 0

$\ln (2 x - 6) = 0$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

{log}_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

e^{0} = 2 x - 6

$e^{0} = 2 x - 6$

Step 1.2.5

Solve for

x

$x$ .

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Step 1.2.5.1

Rewrite the equation as

2 x - 6 = e^{0}

$2 x - 6 = e^{0}$ .

2 x - 6 = e^{0}

$2 x - 6 = e^{0}$

Step 1.2.5.2

Anything raised to

0

$0$ is

1

$1$ .

2 x - 6 = 1

$2 x - 6 = 1$

Step 1.2.5.3

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 1.2.5.3.1

Add

6

$6$ to both sides of the equation.

2 x = 1 + 6

$2 x = 1 + 6$

Step 1.2.5.3.2

Add

1

$1$ and

6

$6$ .

2 x = 7

$2 x = 7$

2 x = 7

$2 x = 7$

Step 1.2.5.4

Divide each term in

2 x = 7

$2 x = 7$ by

2

$2$ and simplify.

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Step 1.2.5.4.1

Divide each term in

2 x = 7

$2 x = 7$ by

2

$2$ .

\frac{2 x}{2} = \frac{7}{2}

$\frac{2 x}{2} = \frac{7}{2}$

Step 1.2.5.4.2

Simplify the left side.

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Step 1.2.5.4.2.1

Cancel the common factor of

2

$2$ .

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Step 1.2.5.4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Step 1.2.5.4.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{7}{2}$

Step 1.3

x-intercept(s) in point form.

x-intercept(s):

$(\frac{7}{2}, 0)$

x-intercept(s):

$(\frac{7}{2}, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in

$0$ for

$x$ and solve for

$y$ .

$y = \ln (2) + \ln ((0) - 3)$

Step 2.2

Solve the equation.

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Step 2.2.1

The natural logarithm of a negative number is undefined.

$y = Undefined$

Step 2.2.2

Remove parentheses.

$y = \ln (2) + \ln ((0) - 3)$

Step 2.2.3

The equation cannot be solved because it is undefined.

$Undefined$

Step 2.3

To find the y-intercept(s), substitute in

$0$ for

$x$ and solve for

$y$ .

y-intercept(s):

$None$

y-intercept(s):

$None$

Step 3

List the intersections.

x-intercept(s):

$(\frac{7}{2}, 0)$

y-intercept(s):

$None$

Step 4