Finite Math Examples

$16 (5^{x}) - x^{2} \cdot 5^{x} = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $5^{x}$ out of $16 \cdot 5^{x} - x^{2} \cdot 5^{x}$ .

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Step 1.1.1

Factor $5^{x}$ out of $16 \cdot 5^{x}$ .

$5^{x} \cdot 16 - x^{2} \cdot 5^{x} = 0$

Step 1.1.2

Factor $5^{x}$ out of $- x^{2} \cdot 5^{x}$ .

$5^{x} \cdot 16 + 5^{x} \cdot (- x^{2}) = 0$

Step 1.1.3

Factor $5^{x}$ out of $5^{x} \cdot 16 + 5^{x} \cdot (- x^{2})$ .

$5^{x} (16 - x^{2}) = 0$

Step 1.2

Rewrite $16$ as $4^{2}$ .

$5^{x} (4^{2} - x^{2}) = 0$

Step 1.3

Factor.

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Step 1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

$5^{x} ((4 + x) (4 - x)) = 0$

Step 1.3.2

Remove unnecessary parentheses.

$5^{x} (4 + x) (4 - x) = 0$

Step 2

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

$5^{x} = 0$

$4 + x = 0$

$4 - x = 0$

Step 3

Set $5^{x}$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $5^{x}$ equal to $0$ .

$5^{x} = 0$

Step 3.2

Solve $5^{x} = 0$ for $x$ .

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Step 3.2.1

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

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

Step 3.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 3.2.3

There is no solution for $5^{x} = 0$

No solution

Step 4

Set $4 + x$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $4 + x$ equal to $0$ .

$4 + x = 0$

Step 4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

Set $4 - x$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $4 - x$ equal to $0$ .

$4 - x = 0$

Step 5.2

Solve $4 - x = 0$ for $x$ .

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Step 5.2.1

Subtract $4$ from both sides of the equation.

$- x = - 4$

Step 5.2.2

Divide each term in $- x = - 4$ by $- 1$ and simplify.

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Step 5.2.2.1

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Step 5.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Divide $- 4$ by $- 1$ .

$x = 4$

Step 6

The final solution is all the values that make $5^{x} (4 + x) (4 - x) = 0$ true.

$x = - 4, 4$