Precalculus Examples

$5^{x^{3} - x} = 1$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (5^{x^{3} - x}) = \ln (1)$

Step 2

Expand $\ln (5^{x^{3} - x})$ by moving $x^{3} - x$ outside the logarithm.

$(x^{3} - x) \ln (5) = \ln (1)$

Step 3

Simplify the left side.

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Step 3.1

Apply the distributive property.

$x^{3} \ln (5) - x \ln (5) = \ln (1)$

Step 4

Simplify the right side.

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Step 4.1

The natural logarithm of $1$ is $0$ .

$x^{3} \ln (5) - x \ln (5) = 0$

Step 5

Factor $x \ln (5)$ out of $x^{3} \ln (5) - x \ln (5)$ .

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Step 5.1

Factor $x \ln (5)$ out of $x^{3} \ln (5)$ .

$x \ln (5) x^{2} - x \ln (5) = 0$

Step 5.2

Factor $x \ln (5)$ out of $- x \ln (5)$ .

$x \ln (5) x^{2} + x \ln (5) \cdot - 1 = 0$

Step 5.3

Factor $x \ln (5)$ out of $x \ln (5) x^{2} + x \ln (5) \cdot - 1$ .

$x \ln (5) (x^{2} - 1) = 0$

Step 6

Rewrite $1$ as $1^{2}$ .

$x \ln (5) (x^{2} - 1^{2}) = 0$

Step 7

Factor.

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Step 7.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$x \ln (5) ((x + 1) (x - 1)) = 0$

Step 7.2

Remove unnecessary parentheses.

$x \ln (5) (x + 1) (x - 1) = 0$

Step 8

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 9

Set $x$ equal to $0$ .

$x = 0$

Step 10

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 10.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 10.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 11.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 11.2

Add $1$ to both sides of the equation.

$x = 1$

Step 12

The final solution is all the values that make $x \ln (5) (x + 1) (x - 1) = 0$ true.

$x = 0, - 1, 1$