Finite Math Examples

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

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

Step 1

Set the argument in

$\ln (x - e^{6} x)$ greater than

$0$ to find where the expression is defined.

$x - e^{6} x > 0$

Step 2

Solve for

$x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor

$x$ out of

$x - e^{6} x$ .

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Step 2.1.1.1

Raise

$x$ to the power of

$1$ .

$x - e^{6} x > 0$

Step 2.1.1.2

Factor

$x$ out of

$x^{1}$ .

$x \cdot 1 - e^{6} x > 0$

Step 2.1.1.3

Factor

$x$ out of

$- e^{6} x$ .

$x \cdot 1 + x (- e^{6}) > 0$

Step 2.1.1.4

Factor

$x$ out of

$x \cdot 1 + x (- e^{6})$ .

$x (1 - e^{6}) > 0$

Step 2.1.2

Rewrite

$1$ as

$1^{3}$ .

$x (1^{3} - e^{6}) > 0$

Step 2.1.3

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$x (1^{3} - {(e^{2})}^{3}) > 0$

Step 2.1.4

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$x ((1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) > 0$

Step 2.1.5

Factor.

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Step 2.1.5.1

Simplify.

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Step 2.1.5.1.1

Rewrite

$1$ as

$1^{2}$ .

$x ((1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) > 0$

Step 2.1.5.1.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$x ((1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) > 0$

Step 2.1.5.1.3

Multiply

$e^{2}$ by

$1$ .

$x ((1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})) > 0$

Step 2.1.5.2

Remove unnecessary parentheses.

$x (1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2}) > 0$

Step 2.1.6

One to any power is one.

$x (1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2}) > 0$

Step 2.1.7

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 2.1.7.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$x (1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2}) > 0$

Step 2.1.7.2

Multiply

$2$ by

$2$ .

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) > 0$

Step 2.2

Divide each term in

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) > 0$ by

$1 - e^{6}$ and simplify.

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Step 2.2.1

Divide each term in

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) > 0$ by

$1 - e^{6}$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1 - e^{6}} < \frac{0}{1 - e^{6}}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Simplify the denominator.

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Step 2.2.2.1.1

Rewrite

$1$ as

$1^{3}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1^{3} - e^{6}} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.2

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1^{3} - {(e^{2})}^{3}} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.4

Simplify.

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Step 2.2.2.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.4.3

Multiply

$e^{2}$ by

$1$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.5

Simplify each term.

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Step 2.2.2.1.5.1

One to any power is one.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.5.2

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 2.2.2.1.5.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.1.5.2.2

Multiply

$2$ by

$2$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.2

Reduce the expression by cancelling the common factors.

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Step 2.2.2.2.1

Cancel the common factor of

$1 + e$ .

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Step 2.2.2.2.1.1

Cancel the common factor.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.2.1.2

Rewrite the expression.

$\frac{(x (1 - e)) (1 + e^{2} + e^{4})}{(1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.2.2

Cancel the common factor of

$1 - e$ .

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Step 2.2.2.2.2.1

Cancel the common factor.

$\frac{x (1 - e) (1 + e^{2} + e^{4})}{(1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 2.2.2.2.2.2

Rewrite the expression.

$\frac{(x) (1 + e^{2} + e^{4})}{1 + e^{2} + e^{4}} < \frac{0}{1 - e^{6}}$

Step 2.2.2.2.3

Cancel the common factor of

$1 + e^{2} + e^{4}$ .

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Step 2.2.2.2.3.1

Cancel the common factor.

$\frac{x (1 + e^{2} + e^{4})}{1 + e^{2} + e^{4}} < \frac{0}{1 - e^{6}}$

Step 2.2.2.2.3.2

Divide

$x$ by

$1$ .

$x < \frac{0}{1 - e^{6}}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Simplify the denominator.

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Step 2.2.3.1.1

Rewrite

$1$ as

$1^{3}$ .

$x < \frac{0}{1^{3} - e^{6}}$

Step 2.2.3.1.2

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$x < \frac{0}{1^{3} - {(e^{2})}^{3}}$

Step 2.2.3.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$x < \frac{0}{(1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 2.2.3.1.4

Simplify.

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Step 2.2.3.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$x < \frac{0}{(1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 2.2.3.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$x < \frac{0}{(1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 2.2.3.1.4.3

Multiply

$e^{2}$ by

$1$ .

$x < \frac{0}{(1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})}$

Step 2.2.3.1.5

Simplify each term.

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Step 2.2.3.1.5.1

One to any power is one.

$x < \frac{0}{(1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2})}$

Step 2.2.3.1.5.2

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 2.2.3.1.5.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$x < \frac{0}{(1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2})}$

Step 2.2.3.1.5.2.2

Multiply

$2$ by

$2$ .

$x < \frac{0}{(1 + e) (1 - e) (1 + e^{2} + e^{4})}$

Step 2.2.3.2

Divide

$0$ by

$(1 + e) (1 - e) (1 + e^{2} + e^{4})$ .

$x < 0$

Step 3

Set the argument in

$\ln (\ln (x - e^{6} x))$ greater than

$0$ to find where the expression is defined.

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

Step 4

Solve for

$x$ .

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Step 4.1

Convert the inequality to an equality.

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

Step 4.2

Solve the equation.

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Step 4.2.1

To solve for

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

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

Step 4.2.2

Rewrite

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

$x$ and

$b$ are positive real numbers and

$b \neq 1$ , then

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

$b^{y} = x$ .

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

Step 4.2.3

Solve for

$x$ .

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Step 4.2.3.1

Rewrite the equation as

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

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

Step 4.2.3.2

Anything raised to

$0$ is

$1$ .

$x - e^{6} x = 1$

Step 4.2.3.3

Factor the left side of the equation.

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Step 4.2.3.3.1

Factor

$x$ out of

$x - e^{6} x$ .

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Step 4.2.3.3.1.1

Raise

$x$ to the power of

$1$ .

$x - e^{6} x = 1$

Step 4.2.3.3.1.2

Factor

$x$ out of

$x^{1}$ .

$x \cdot 1 - e^{6} x = 1$

Step 4.2.3.3.1.3

Factor

$x$ out of

$- e^{6} x$ .

$x \cdot 1 + x (- e^{6}) = 1$

Step 4.2.3.3.1.4

Factor

$x$ out of

$x \cdot 1 + x (- e^{6})$ .

$x (1 - e^{6}) = 1$

Step 4.2.3.3.2

Rewrite

$1$ as

$1^{3}$ .

$x (1^{3} - e^{6}) = 1$

Step 4.2.3.3.3

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$x (1^{3} - {(e^{2})}^{3}) = 1$

Step 4.2.3.3.4

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$x ((1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) = 1$

Step 4.2.3.3.5

Factor.

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Step 4.2.3.3.5.1

Simplify.

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Step 4.2.3.3.5.1.1

Rewrite

$1$ as

$1^{2}$ .

$x ((1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) = 1$

Step 4.2.3.3.5.1.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$x ((1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) = 1$

Step 4.2.3.3.5.1.3

Multiply

$e^{2}$ by

$1$ .

$x ((1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})) = 1$

Step 4.2.3.3.5.2

Remove unnecessary parentheses.

$x (1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2}) = 1$

Step 4.2.3.3.6

One to any power is one.

$x (1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2}) = 1$

Step 4.2.3.3.7

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 4.2.3.3.7.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$x (1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2}) = 1$

Step 4.2.3.3.7.2

Multiply

$2$ by

$2$ .

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) = 1$

Step 4.2.3.4

Divide each term in

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) = 1$ by

$1 - e^{6}$ and simplify.

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Step 4.2.3.4.1

Divide each term in

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) = 1$ by

$1 - e^{6}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1 - e^{6}} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2

Simplify the left side.

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Step 4.2.3.4.2.1

Simplify the denominator.

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Step 4.2.3.4.2.1.1

Rewrite

$1$ as

$1^{3}$ .

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

Step 4.2.3.4.2.1.2

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

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

Step 4.2.3.4.2.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.1.4

Simplify.

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Step 4.2.3.4.2.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.1.4.3

Multiply

$e^{2}$ by

$1$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.1.5

Simplify each term.

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Step 4.2.3.4.2.1.5.1

One to any power is one.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.1.5.2

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 4.2.3.4.2.1.5.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.1.5.2.2

Multiply

$2$ by

$2$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{4})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.2

Reduce the expression by cancelling the common factors.

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Step 4.2.3.4.2.2.1

Cancel the common factor of

$1 + e$ .

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Step 4.2.3.4.2.2.1.1

Cancel the common factor.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{4})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.2.1.2

Rewrite the expression.

$\frac{(x (1 - e)) (1 + e^{2} + e^{4})}{(1 - e) (1 + e^{2} + e^{4})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.2.2

Cancel the common factor of

$1 - e$ .

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Step 4.2.3.4.2.2.2.1

Cancel the common factor.

$\frac{x (1 - e) (1 + e^{2} + e^{4})}{(1 - e) (1 + e^{2} + e^{4})} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.2.2.2

Rewrite the expression.

$\frac{(x) (1 + e^{2} + e^{4})}{1 + e^{2} + e^{4}} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.2.3

Cancel the common factor of

$1 + e^{2} + e^{4}$ .

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Step 4.2.3.4.2.2.3.1

Cancel the common factor.

$\frac{x (1 + e^{2} + e^{4})}{1 + e^{2} + e^{4}} = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.2.2.3.2

Divide

$x$ by

$1$ .

$x = \frac{1}{1 - e^{6}}$

Step 4.2.3.4.3

Simplify the right side.

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Step 4.2.3.4.3.1

Simplify the denominator.

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Step 4.2.3.4.3.1.1

Rewrite

$1$ as

$1^{3}$ .

$x = \frac{1}{1^{3} - e^{6}}$

Step 4.2.3.4.3.1.2

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$x = \frac{1}{1^{3} - {(e^{2})}^{3}}$

Step 4.2.3.4.3.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$x = \frac{1}{(1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 4.2.3.4.3.1.4

Simplify.

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Step 4.2.3.4.3.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$x = \frac{1}{(1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 4.2.3.4.3.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$x = \frac{1}{(1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 4.2.3.4.3.1.4.3

Multiply

$e^{2}$ by

$1$ .

$x = \frac{1}{(1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})}$

Step 4.2.3.4.3.1.5

Simplify each term.

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Step 4.2.3.4.3.1.5.1

One to any power is one.

$x = \frac{1}{(1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2})}$

Step 4.2.3.4.3.1.5.2

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 4.2.3.4.3.1.5.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{1}{(1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2})}$

Step 4.2.3.4.3.1.5.2.2

Multiply

$2$ by

$2$ .

$x = \frac{1}{(1 + e) (1 - e) (1 + e^{2} + e^{4})}$

Step 4.3

Find the domain of

$\ln (x - e^{6} x)$ .

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Step 4.3.1

Set the argument in

$\ln (x - e^{6} x)$ greater than

$0$ to find where the expression is defined.

$x - e^{6} x > 0$

Step 4.3.2

Solve for

$x$ .

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Step 4.3.2.1

Factor the left side of the equation.

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Step 4.3.2.1.1

Factor

$x$ out of

$x - e^{6} x$ .

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Step 4.3.2.1.1.1

Raise

$x$ to the power of

$1$ .

$x - e^{6} x > 0$

Step 4.3.2.1.1.2

Factor

$x$ out of

$x^{1}$ .

$x \cdot 1 - e^{6} x > 0$

Step 4.3.2.1.1.3

Factor

$x$ out of

$- e^{6} x$ .

$x \cdot 1 + x (- e^{6}) > 0$

Step 4.3.2.1.1.4

Factor

$x$ out of

$x \cdot 1 + x (- e^{6})$ .

$x (1 - e^{6}) > 0$

Step 4.3.2.1.2

Rewrite

$1$ as

$1^{3}$ .

$x (1^{3} - e^{6}) > 0$

Step 4.3.2.1.3

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$x (1^{3} - {(e^{2})}^{3}) > 0$

Step 4.3.2.1.4

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$x ((1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) > 0$

Step 4.3.2.1.5

Factor.

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Step 4.3.2.1.5.1

Simplify.

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Step 4.3.2.1.5.1.1

Rewrite

$1$ as

$1^{2}$ .

$x ((1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) > 0$

Step 4.3.2.1.5.1.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$x ((1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})) > 0$

Step 4.3.2.1.5.1.3

Multiply

$e^{2}$ by

$1$ .

$x ((1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})) > 0$

Step 4.3.2.1.5.2

Remove unnecessary parentheses.

$x (1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2}) > 0$

Step 4.3.2.1.6

One to any power is one.

$x (1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2}) > 0$

Step 4.3.2.1.7

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 4.3.2.1.7.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$x (1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2}) > 0$

Step 4.3.2.1.7.2

Multiply

$2$ by

$2$ .

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) > 0$

Step 4.3.2.2

Divide each term in

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) > 0$ by

$1 - e^{6}$ and simplify.

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Step 4.3.2.2.1

Divide each term in

$x (1 + e) (1 - e) (1 + e^{2} + e^{4}) > 0$ by

$1 - e^{6}$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1 - e^{6}} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2

Simplify the left side.

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Step 4.3.2.2.2.1

Simplify the denominator.

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Step 4.3.2.2.2.1.1

Rewrite

$1$ as

$1^{3}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1^{3} - e^{6}} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.2

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{1^{3} - {(e^{2})}^{3}} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.4

Simplify.

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Step 4.3.2.2.2.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.4.3

Multiply

$e^{2}$ by

$1$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.5

Simplify each term.

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Step 4.3.2.2.2.1.5.1

One to any power is one.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.5.2

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 4.3.2.2.2.1.5.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.1.5.2.2

Multiply

$2$ by

$2$ .

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.2

Reduce the expression by cancelling the common factors.

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Step 4.3.2.2.2.2.1

Cancel the common factor of

$1 + e$ .

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Step 4.3.2.2.2.2.1.1

Cancel the common factor.

$\frac{x (1 + e) (1 - e) (1 + e^{2} + e^{4})}{(1 + e) (1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.2.1.2

Rewrite the expression.

$\frac{(x (1 - e)) (1 + e^{2} + e^{4})}{(1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.2.2

Cancel the common factor of

$1 - e$ .

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Step 4.3.2.2.2.2.2.1

Cancel the common factor.

$\frac{x (1 - e) (1 + e^{2} + e^{4})}{(1 - e) (1 + e^{2} + e^{4})} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.2.2.2

Rewrite the expression.

$\frac{(x) (1 + e^{2} + e^{4})}{1 + e^{2} + e^{4}} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.2.3

Cancel the common factor of

$1 + e^{2} + e^{4}$ .

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Step 4.3.2.2.2.2.3.1

Cancel the common factor.

$\frac{x (1 + e^{2} + e^{4})}{1 + e^{2} + e^{4}} < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.2.2.3.2

Divide

$x$ by

$1$ .

$x < \frac{0}{1 - e^{6}}$

Step 4.3.2.2.3

Simplify the right side.

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Step 4.3.2.2.3.1

Simplify the denominator.

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Step 4.3.2.2.3.1.1

Rewrite

$1$ as

$1^{3}$ .

$x < \frac{0}{1^{3} - e^{6}}$

Step 4.3.2.2.3.1.2

Rewrite

$e^{6}$ as

${(e^{2})}^{3}$ .

$x < \frac{0}{1^{3} - {(e^{2})}^{3}}$

Step 4.3.2.2.3.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula,

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

$a = 1$ and

$b = e^{2}$ .

$x < \frac{0}{(1 - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 4.3.2.2.3.1.4

Simplify.

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Step 4.3.2.2.3.1.4.1

Rewrite

$1$ as

$1^{2}$ .

$x < \frac{0}{(1^{2} - e^{2}) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 4.3.2.2.3.1.4.2

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 1$ and

$b = e$ .

$x < \frac{0}{(1 + e) (1 - e) (1^{2} + 1 e^{2} + {(e^{2})}^{2})}$

Step 4.3.2.2.3.1.4.3

Multiply

$e^{2}$ by

$1$ .

$x < \frac{0}{(1 + e) (1 - e) (1^{2} + e^{2} + {(e^{2})}^{2})}$

Step 4.3.2.2.3.1.5

Simplify each term.

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Step 4.3.2.2.3.1.5.1

One to any power is one.

$x < \frac{0}{(1 + e) (1 - e) (1 + e^{2} + {(e^{2})}^{2})}$

Step 4.3.2.2.3.1.5.2

Multiply the exponents in

${(e^{2})}^{2}$ .

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Step 4.3.2.2.3.1.5.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$x < \frac{0}{(1 + e) (1 - e) (1 + e^{2} + e^{2 \cdot 2})}$

Step 4.3.2.2.3.1.5.2.2

Multiply

$2$ by

$2$ .

$x < \frac{0}{(1 + e) (1 - e) (1 + e^{2} + e^{4})}$

Step 4.3.2.2.3.2

Divide

$0$ by

$(1 + e) (1 - e) (1 + e^{2} + e^{4})$ .

$x < 0$

Step 4.3.3

The domain is all values of

$x$ that make the expression defined.

$(- \infty, 0)$

Step 4.4

The solution consists of all of the true intervals.

$x < \frac{1}{(1 + e) (1 - e) (1 + e^{2} + e^{4})}$

Step 5

The domain is all values of

$x$ that make the expression defined.

Interval Notation:

$(- \infty, \frac{1}{(1 + e) (1 - e) (1 + e^{2} + e^{4})})$

Set-Builder Notation:

${x | x < \frac{1}{(1 + e) (1 - e) (1 + e^{2} + e^{4})}}$

Step 6