Calculus Examples

$f (x) = \frac{3 x}{2^{x}}$ , $x = 2$

Step 1

Find the corresponding $y$ -value to $x = 2$ .

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Step 1.1

Substitute $2$ in for $x$ .

$y = \frac{3 (2)}{2^{2}}$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = \frac{3 (2)}{2^{2}}$

Step 1.2.2

Cancel the common factor of $2$ and $2^{2}$ .

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Step 1.2.2.1

Factor $2$ out of $3 (2)$ .

$y = \frac{2 \cdot 3}{2^{2}}$

Step 1.2.2.2

Cancel the common factors.

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Step 1.2.2.2.1

Factor $2$ out of $2^{2}$ .

$y = \frac{2 \cdot 3}{2 \cdot 2}$

Step 1.2.2.2.2

Cancel the common factor.

$y = \frac{2 \cdot 3}{2 \cdot 2}$

Step 1.2.2.2.3

Rewrite the expression.

$y = \frac{3}{2}$

Step 2

Find the first derivative and evaluate at $x = 2$ and $y = \frac{3}{2}$ to find the slope of the tangent line.

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Step 2.1

Since $3$ is constant with respect to $x$ , the derivative of $\frac{3 x}{2^{x}}$ with respect to $x$ is $3 \frac{d}{d x} [\frac{x}{2^{x}}]$ .

$3 \frac{d}{d x} [\frac{x}{2^{x}}]$

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = 2^{x}$ .

$3 \frac{2^{x} \frac{d}{d x} [x] - x \frac{d}{d x} [2^{x}]}{{(2^{x})}^{2}}$

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

Multiply the exponents in ${(2^{x})}^{2}$ .

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Step 2.3.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$3 \frac{2^{x} \frac{d}{d x} [x] - x \frac{d}{d x} [2^{x}]}{2^{x \cdot 2}}$

Step 2.3.1.2

Move $2$ to the left of $x$ .

$3 \frac{2^{x} \frac{d}{d x} [x] - x \frac{d}{d x} [2^{x}]}{2^{2 x}}$

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$3 \frac{2^{x} \cdot 1 - x \frac{d}{d x} [2^{x}]}{2^{2 x}}$

Step 2.3.3

Multiply $2^{x}$ by $1$ .

$3 \frac{2^{x} - x \frac{d}{d x} [2^{x}]}{2^{2 x}}$

Step 2.4

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $2$ .

$3 \frac{2^{x} - x (2^{x} \ln (2))}{2^{2 x}}$

Step 2.5

Combine $3$ and $\frac{2^{x} - x (2^{x} \ln (2))}{2^{2 x}}$ .

$\frac{3 (2^{x} - x (2^{x} \ln (2)))}{2^{2 x}}$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$\frac{3 \cdot 2^{x} + 3 (- x (2^{x} \ln (2)))}{2^{2 x}}$

Step 2.6.2

Simplify the numerator.

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Step 2.6.2.1

Simplify each term.

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Step 2.6.2.1.1

Multiply $3 (- x (2^{x} \ln (2)))$ .

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Step 2.6.2.1.1.1

Multiply $- 1$ by $3$ .

$\frac{3 \cdot 2^{x} - 3 (x (2^{x} \ln (2)))}{2^{2 x}}$

Step 2.6.2.1.1.2

Simplify $- 3 \ln (2)$ by moving $3$ inside the logarithm.

$\frac{3 \cdot 2^{x} + x (2^{x} (- \ln (2^{3})))}{2^{2 x}}$

Step 2.6.2.1.2

Rewrite using the commutative property of multiplication.

$\frac{3 \cdot 2^{x} - \ln (2^{3}) x \cdot 2^{x}}{2^{2 x}}$

Step 2.6.2.1.3

Raise $2$ to the power of $3$ .

$\frac{3 \cdot 2^{x} - \ln (8) x \cdot 2^{x}}{2^{2 x}}$

Step 2.6.2.2

Reorder factors in $3 \cdot 2^{x} - \ln (8) x \cdot 2^{x}$ .

$\frac{3 \cdot 2^{x} - x \cdot 2^{x} \ln (8)}{2^{2 x}}$

Step 2.6.3

Reorder terms.

$\frac{- 2^{x} x \ln (8) + 3 \cdot 2^{x}}{2^{2 x}}$

Step 2.6.4

Factor $2^{x}$ out of $- 2^{x} x \ln (8) + 3 \cdot 2^{x}$ .

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Step 2.6.4.1

Factor $2^{x}$ out of $- 2^{x} x \ln (8)$ .

$\frac{2^{x} (- x \ln (8)) + 3 \cdot 2^{x}}{2^{2 x}}$

Step 2.6.4.2

Factor $2^{x}$ out of $3 \cdot 2^{x}$ .

$\frac{2^{x} (- x \ln (8)) + 2^{x} \cdot 3}{2^{2 x}}$

Step 2.6.4.3

Factor $2^{x}$ out of $2^{x} (- x \ln (8)) + 2^{x} \cdot 3$ .

$\frac{2^{x} (- x \ln (8) + 3)}{2^{2 x}}$

Step 2.6.5

Cancel the common factor of $2^{x}$ and $2^{2 x}$ .

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Step 2.6.5.1

Factor $2^{2 x}$ out of $2^{x} (- x \ln (8) + 3)$ .

$\frac{2^{2 x} (2^{- x} (- x \ln (8) + 3))}{2^{2 x}}$

Step 2.6.5.2

Cancel the common factors.

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Step 2.6.5.2.1

Multiply by $1$ .

$\frac{2^{2 x} (2^{- x} (- x \ln (8) + 3))}{2^{2 x} \cdot 1}$

Step 2.6.5.2.2

Cancel the common factor.

$\frac{2^{2 x} (2^{- x} (- x \ln (8) + 3))}{2^{2 x} \cdot 1}$

Step 2.6.5.2.3

Rewrite the expression.

$\frac{2^{- x} (- x \ln (8) + 3)}{1}$

Step 2.6.5.2.4

Divide $2^{- x} (- x \ln (8) + 3)$ by $1$ .

$2^{- x} (- x \ln (8) + 3)$

Step 2.6.6

Apply the distributive property.

$2^{- x} (- x \ln (8)) + 2^{- x} \cdot 3$

Step 2.6.7

Rewrite using the commutative property of multiplication.

$- \ln (8) \cdot 2^{- x} x + 2^{- x} \cdot 3$

Step 2.6.8

Move $3$ to the left of $2^{- x}$ .

$- \ln (8) \cdot 2^{- x} x + 3 \cdot 2^{- x}$

Step 2.6.9

Reorder factors in $- \ln (8) \cdot 2^{- x} x + 3 \cdot 2^{- x}$ .

$- x \cdot 2^{- x} \ln (8) + 3 \cdot 2^{- x}$

Step 2.7

Evaluate the derivative at $x = 2$ .

$- 2 \cdot (2^{- (2)} \ln (8)) + 3 \cdot 2^{- (2)}$

Step 2.8

Simplify each term.

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Step 2.8.1

Multiply $- 1$ by $2$ .

$- 2 \cdot (2^{- 2} \ln (8)) + 3 \cdot 2^{- (2)}$

Step 2.8.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- 2 \cdot (\frac{1}{2^{2}} \ln (8)) + 3 \cdot 2^{- (2)}$

Step 2.8.3

Raise $2$ to the power of $2$ .

$- 2 \cdot (\frac{1}{4} \ln (8)) + 3 \cdot 2^{- (2)}$

Step 2.8.4

Simplify $\frac{1}{4} \ln (8)$ by moving $\frac{1}{4}$ inside the logarithm.

$- 2 \cdot \ln (8^{\frac{1}{4}}) + 3 \cdot 2^{- (2)}$

Step 2.8.5

Simplify $- 2 \ln (8^{\frac{1}{4}})$ by moving $2$ inside the logarithm.

$- \ln ({(8^{\frac{1}{4}})}^{2}) + 3 \cdot 2^{- (2)}$

Step 2.8.6

Multiply the exponents in ${(8^{\frac{1}{4}})}^{2}$ .

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Step 2.8.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \ln (8^{\frac{1}{4} \cdot 2}) + 3 \cdot 2^{- (2)}$

Step 2.8.6.2

Cancel the common factor of $2$ .

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Step 2.8.6.2.1

Factor $2$ out of $4$ .

$- \ln (8^{\frac{1}{2 (2)} \cdot 2}) + 3 \cdot 2^{- (2)}$

Step 2.8.6.2.2

Cancel the common factor.

$- \ln (8^{\frac{1}{2 \cdot 2} \cdot 2}) + 3 \cdot 2^{- (2)}$

Step 2.8.6.2.3

Rewrite the expression.

$- \ln (8^{\frac{1}{2}}) + 3 \cdot 2^{- (2)}$

Step 2.8.7

Multiply $- 1$ by $2$ .

$- \ln (8^{\frac{1}{2}}) + 3 \cdot 2^{- 2}$

Step 2.8.8

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \ln (8^{\frac{1}{2}}) + 3 \cdot \frac{1}{2^{2}}$

Step 2.8.9

Raise $2$ to the power of $2$ .

$- \ln (8^{\frac{1}{2}}) + 3 \cdot \frac{1}{4}$

Step 2.8.10

Combine $3$ and $\frac{1}{4}$ .

$- \ln (8^{\frac{1}{2}}) + \frac{3}{4}$

Step 3

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 3.1

Use the slope $- \ln (8^{\frac{1}{2}}) + \frac{3}{4}$ and a given point $(2, \frac{3}{2})$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (\frac{3}{2}) = (- \ln (8^{\frac{1}{2}}) + \frac{3}{4}) \cdot (x - (2))$

Step 3.2

Simplify the equation and keep it in point-slope form.

$y - \frac{3}{2} = (- \ln (8^{\frac{1}{2}}) + \frac{3}{4}) \cdot (x - 2)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $(- \ln (8^{\frac{1}{2}}) + \frac{3}{4}) \cdot (x - 2)$ .

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Step 3.3.1.1

Rewrite.

$y - \frac{3}{2} = 0 + 0 + (- \ln (8^{\frac{1}{2}}) + \frac{3}{4}) \cdot (x - 2)$

Step 3.3.1.2

Simplify by adding zeros.

$y - \frac{3}{2} = (- \ln (8^{\frac{1}{2}}) + \frac{3}{4}) \cdot (x - 2)$

Step 3.3.1.3

Expand $(- \ln (8^{\frac{1}{2}}) + \frac{3}{4}) (x - 2)$ using the FOIL Method.

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Step 3.3.1.3.1

Apply the distributive property.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) (x - 2) + \frac{3}{4} (x - 2)$

Step 3.3.1.3.2

Apply the distributive property.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x - \ln (8^{\frac{1}{2}}) \cdot - 2 + \frac{3}{4} (x - 2)$

Step 3.3.1.3.3

Apply the distributive property.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x - \ln (8^{\frac{1}{2}}) \cdot - 2 + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4

Simplify terms.

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Step 3.3.1.4.1

Simplify each term.

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Step 3.3.1.4.1.1

Multiply $- \ln (8^{\frac{1}{2}}) \cdot - 2$ .

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Step 3.3.1.4.1.1.1

Multiply $- 2$ by $- 1$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + 2 \ln (8^{\frac{1}{2}}) + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.1.2

Simplify $2 \ln (8^{\frac{1}{2}})$ by moving $2$ inside the logarithm.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln ({(8^{\frac{1}{2}})}^{2}) + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.2

Multiply the exponents in ${(8^{\frac{1}{2}})}^{2}$ .

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Step 3.3.1.4.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8^{\frac{1}{2} \cdot 2}) + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.2.2

Cancel the common factor of $2$ .

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Step 3.3.1.4.1.2.2.1

Cancel the common factor.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8^{\frac{1}{2} \cdot 2}) + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.2.2.2

Rewrite the expression.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8^{1}) + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.3

Evaluate the exponent.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3}{4} x + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.4

Combine $\frac{3}{4}$ and $x$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{3}{4} \cdot - 2$

Step 3.3.1.4.1.5

Cancel the common factor of $2$ .

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Step 3.3.1.4.1.5.1

Factor $2$ out of $4$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{3}{2 (2)} \cdot - 2$

Step 3.3.1.4.1.5.2

Factor $2$ out of $- 2$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{3}{2 \cdot 2} \cdot (2 \cdot - 1)$

Step 3.3.1.4.1.5.3

Cancel the common factor.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{3}{2 \cdot 2} \cdot (2 \cdot - 1)$

Step 3.3.1.4.1.5.4

Rewrite the expression.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{3}{2} \cdot - 1$

Step 3.3.1.4.1.6

Combine $\frac{3}{2}$ and $- 1$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{3 \cdot - 1}{2}$

Step 3.3.1.4.1.7

Multiply $3$ by $- 1$ .

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} + \frac{- 3}{2}$

Step 3.3.1.4.1.8

Move the negative in front of the fraction.

$y - \frac{3}{2} = - \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} - \frac{3}{2}$

Step 3.3.1.4.2

Reorder factors in $- \ln (8^{\frac{1}{2}}) x + \ln (8) + \frac{3 x}{4} - \frac{3}{2}$ .

$y - \frac{3}{2} = - x \ln (8^{\frac{1}{2}}) + \ln (8) + \frac{3 x}{4} - \frac{3}{2}$

Step 3.3.2

Multiply each term in $y - \frac{3}{2} = - x \ln (8^{\frac{1}{2}}) + \ln (8) + \frac{3 x}{4} - \frac{3}{2}$ by $4$ to eliminate the fractions.

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Step 3.3.2.1

Multiply each term in $y - \frac{3}{2} = - x \ln (8^{\frac{1}{2}}) + \ln (8) + \frac{3 x}{4} - \frac{3}{2}$ by $4$ .

$y \cdot 4 - \frac{3}{2} \cdot 4 = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Simplify each term.

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Step 3.3.2.2.1.1

Move $4$ to the left of $y$ .

$4 \cdot y - \frac{3}{2} \cdot 4 = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.2.1.2

Cancel the common factor of $2$ .

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Step 3.3.2.2.1.2.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$4 y + \frac{- 3}{2} \cdot 4 = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.2.1.2.2

Factor $2$ out of $4$ .

$4 y + \frac{- 3}{2} \cdot (2 (2)) = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.2.1.2.3

Cancel the common factor.

$4 y + \frac{- 3}{2} \cdot (2 \cdot 2) = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.2.1.2.4

Rewrite the expression.

$4 y - 3 \cdot 2 = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.2.1.3

Multiply $- 3$ by $2$ .

$4 y - 6 = - x \ln (8^{\frac{1}{2}}) \cdot 4 + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Simplify each term.

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Step 3.3.2.3.1.1

Multiply $4$ by $- 1$ .

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + \ln (8) \cdot 4 + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.3.1.2

Move $4$ to the left of $\ln (8)$ .

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \cdot \ln (8) + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.3.1.3

Cancel the common factor of $4$ .

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Step 3.3.2.3.1.3.1

Cancel the common factor.

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + \frac{3 x}{4} \cdot 4 - \frac{3}{2} \cdot 4$

Step 3.3.2.3.1.3.2

Rewrite the expression.

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x - \frac{3}{2} \cdot 4$

Step 3.3.2.3.1.4

Cancel the common factor of $2$ .

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Step 3.3.2.3.1.4.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x + \frac{- 3}{2} \cdot 4$

Step 3.3.2.3.1.4.2

Factor $2$ out of $4$ .

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x + \frac{- 3}{2} \cdot (2 (2))$

Step 3.3.2.3.1.4.3

Cancel the common factor.

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x + \frac{- 3}{2} \cdot (2 \cdot 2)$

Step 3.3.2.3.1.4.4

Rewrite the expression.

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x - 3 \cdot 2$

Step 3.3.2.3.1.5

Multiply $- 3$ by $2$ .

$4 y - 6 = - 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x - 6$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify $- 4 x \ln (8^{\frac{1}{2}}) + 4 \ln (8) + 3 x - 6$ .

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Step 3.3.3.1.1

Simplify each term.

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Step 3.3.3.1.1.1

Simplify $- 4 \ln (8^{\frac{1}{2}})$ by moving $4$ inside the logarithm.

$4 y - 6 = x (- \ln ({(8^{\frac{1}{2}})}^{4})) + 4 \ln (8) + 3 x - 6$

Step 3.3.3.1.1.2

Multiply the exponents in ${(8^{\frac{1}{2}})}^{4}$ .

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Step 3.3.3.1.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 y - 6 = x (- \ln (8^{\frac{1}{2} \cdot 4})) + 4 \ln (8) + 3 x - 6$

Step 3.3.3.1.1.2.2

Cancel the common factor of $2$ .

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Step 3.3.3.1.1.2.2.1

Factor $2$ out of $4$ .

$4 y - 6 = x (- \ln (8^{\frac{1}{2} \cdot (2 (2))})) + 4 \ln (8) + 3 x - 6$

Step 3.3.3.1.1.2.2.2

Cancel the common factor.

$4 y - 6 = x (- \ln (8^{\frac{1}{2} \cdot (2 \cdot 2)})) + 4 \ln (8) + 3 x - 6$

Step 3.3.3.1.1.2.2.3

Rewrite the expression.

$4 y - 6 = x (- \ln (8^{2})) + 4 \ln (8) + 3 x - 6$

Step 3.3.3.1.1.3

Raise $8$ to the power of $2$ .

$4 y - 6 = x (- \ln (64)) + 4 \ln (8) + 3 x - 6$

Step 3.3.3.1.1.4

Simplify $4 \ln (8)$ by moving $4$ inside the logarithm.

$4 y - 6 = x (- \ln (64)) + \ln (8^{4}) + 3 x - 6$

Step 3.3.3.1.1.5

Raise $8$ to the power of $4$ .

$4 y - 6 = x (- \ln (64)) + \ln (4096) + 3 x - 6$

Step 3.3.3.1.2

Reorder factors in $x (- \ln (64)) + \ln (4096) + 3 x - 6$ .

$4 y - 6 = - x \ln (64) + \ln (4096) + 3 x - 6$

Step 3.3.4

Move all the terms containing a logarithm to the left side of the equation.

$x \ln (64) - \ln (4096) = - 4 y + 6 + 3 x - 6$

Step 3.3.5

Combine the opposite terms in $- 4 y + 6 + 3 x - 6$ .

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Step 3.3.5.1

Subtract $6$ from $6$ .

$x \ln (64) - \ln (4096) = - 4 y + 3 x + 0$

Step 3.3.5.2

Add $- 4 y + 3 x$ and $0$ .

$x \ln (64) - \ln (4096) = - 4 y + 3 x$

Step 3.3.6

Rewrite the equation as $- 4 y + 3 x = x \ln (64) - \ln (4096)$ .

$- 4 y + 3 x = x \ln (64) - \ln (4096)$

Step 3.3.7

Subtract $3 x$ from both sides of the equation.

$- 4 y = x \ln (64) - \ln (4096) - 3 x$

Step 3.3.8

Divide each term in $- 4 y = x \ln (64) - \ln (4096) - 3 x$ by $- 4$ and simplify.

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Step 3.3.8.1

Divide each term in $- 4 y = x \ln (64) - \ln (4096) - 3 x$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{x \ln (64)}{- 4} + \frac{- \ln (4096)}{- 4} + \frac{- 3 x}{- 4}$

Step 3.3.8.2

Simplify the left side.

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Step 3.3.8.2.1

Cancel the common factor of $- 4$ .

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Step 3.3.8.2.1.1

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{x \ln (64)}{- 4} + \frac{- \ln (4096)}{- 4} + \frac{- 3 x}{- 4}$

Step 3.3.8.2.1.2

Divide $y$ by $1$ .

$y = \frac{x \ln (64)}{- 4} + \frac{- \ln (4096)}{- 4} + \frac{- 3 x}{- 4}$

Step 3.3.8.3

Simplify the right side.

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Step 3.3.8.3.1

Simplify each term.

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Step 3.3.8.3.1.1

Move the negative in front of the fraction.

$y = - \frac{x \ln (64)}{4} + \frac{- \ln (4096)}{- 4} + \frac{- 3 x}{- 4}$

Step 3.3.8.3.1.2

Dividing two negative values results in a positive value.

$y = - \frac{x \ln (64)}{4} + \frac{\ln (4096)}{4} + \frac{- 3 x}{- 4}$

Step 3.3.8.3.1.3

Dividing two negative values results in a positive value.

$y = - \frac{x \ln (64)}{4} + \frac{\ln (4096)}{4} + \frac{3 x}{4}$

Step 3.3.9

Write in $y = m x + b$ form.

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Step 3.3.9.1

Simplify $- \frac{x \ln (64)}{4} + \frac{\ln (4096)}{4} + \frac{3 x}{4}$ .

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Step 3.3.9.1.1

Simplify each term.

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Step 3.3.9.1.1.1

Rewrite $\ln (4096)$ as $\ln (2^{12})$ .

$y = - \frac{x \ln (64)}{4} + \frac{\ln (2^{12})}{4} + \frac{3 x}{4}$

Step 3.3.9.1.1.2

Expand $\ln (2^{12})$ by moving $12$ outside the logarithm.

$y = - \frac{x \ln (64)}{4} + \frac{12 \ln (2)}{4} + \frac{3 x}{4}$

Step 3.3.9.1.1.3

Cancel the common factor of $12$ and $4$ .

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Step 3.3.9.1.1.3.1

Factor $4$ out of $12 \ln (2)$ .

$y = - \frac{x \ln (64)}{4} + \frac{4 (3 \ln (2))}{4} + \frac{3 x}{4}$

Step 3.3.9.1.1.3.2

Cancel the common factors.

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Step 3.3.9.1.1.3.2.1

Factor $4$ out of $4$ .

$y = - \frac{x \ln (64)}{4} + \frac{4 (3 \ln (2))}{4 (1)} + \frac{3 x}{4}$

Step 3.3.9.1.1.3.2.2

Cancel the common factor.

$y = - \frac{x \ln (64)}{4} + \frac{4 (3 \ln (2))}{4 \cdot 1} + \frac{3 x}{4}$

Step 3.3.9.1.1.3.2.3

Rewrite the expression.

$y = - \frac{x \ln (64)}{4} + \frac{3 \ln (2)}{1} + \frac{3 x}{4}$

Step 3.3.9.1.1.3.2.4

Divide $3 \ln (2)$ by $1$ .

$y = - \frac{x \ln (64)}{4} + 3 \ln (2) + \frac{3 x}{4}$

Step 3.3.9.1.1.4

Simplify $3 \ln (2)$ by moving $3$ inside the logarithm.

$y = - \frac{x \ln (64)}{4} + \ln (2^{3}) + \frac{3 x}{4}$

Step 3.3.9.1.1.5

Raise $2$ to the power of $3$ .

$y = - \frac{x \ln (64)}{4} + \ln (8) + \frac{3 x}{4}$

Step 3.3.9.1.2

Move $\ln (8)$ .

$y = - \frac{x \ln (64)}{4} + \frac{3 x}{4} + \ln (8)$

Step 3.3.9.2

Combine the numerators over the common denominator.

$y = \frac{- x \ln (64) + 3 x}{4} + \ln (8)$

Step 3.3.9.3

Simplify the numerator.

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Step 3.3.9.3.1

Factor $x$ out of $- x \ln (64) + 3 x$ .

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Step 3.3.9.3.1.1

Factor $x$ out of $- x \ln (64)$ .

$y = \frac{x (- 1 \ln (64)) + 3 x}{4} + \ln (8)$

Step 3.3.9.3.1.2

Factor $x$ out of $3 x$ .

$y = \frac{x (- 1 \ln (64)) + x \cdot 3}{4} + \ln (8)$

Step 3.3.9.3.1.3

Factor $x$ out of $x (- 1 \ln (64)) + x \cdot 3$ .

$y = \frac{x (- 1 \ln (64) + 3)}{4} + \ln (8)$

Step 3.3.9.3.2

Rewrite $- 1 \ln (64)$ as $- \ln (64)$ .

$y = \frac{x (- \ln (64) + 3)}{4} + \ln (8)$

Step 3.3.9.4

Factor $- 1$ out of $- \ln (64)$ .

$y = \frac{x (- (\ln (64)) + 3)}{4} + \ln (8)$

Step 3.3.9.5

Rewrite $3$ as $- 1 (- 3)$ .

$y = \frac{x (- (\ln (64)) - 1 (- 3))}{4} + \ln (8)$

Step 3.3.9.6

Factor $- 1$ out of $- (\ln (64)) - 1 (- 3)$ .

$y = \frac{x (- (\ln (64) - 3))}{4} + \ln (8)$

Step 3.3.9.7

Rewrite $- (\ln (64) - 3)$ as $- 1 (\ln (64) - 3)$ .

$y = \frac{x (- 1 (\ln (64) - 3))}{4} + \ln (8)$

Step 3.3.9.8

Move the negative in front of the fraction.

$y = - \frac{x (\ln (64) - 3)}{4} + \ln (8)$

Step 3.3.9.9

Reorder terms.

$y = - (\frac{\ln (64) - 3}{4} x) + \ln (8)$

Step 3.3.9.10

Remove parentheses.

$y = - \frac{\ln (64) - 3}{4} x + \ln (8)$

Step 4