Pre-Algebra Examples

$9^{5 x - x^{2}} = 9^{4}$

Step 1

Create equivalent expressions in the equation that all have equal bases.

$9^{5 x - x^{2}} = 9^{4}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$5 x - x^{2} = 4$

Step 3

Solve for $x$ .

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Step 3.1

Subtract $4$ from both sides of the equation.

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

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $- 1$ out of $5 x - x^{2} - 4$ .

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Step 3.2.1.1

Reorder $5 x$ and $- x^{2}$ .

$- x^{2} + 5 x - 4 = 0$

Step 3.2.1.2

Factor $- 1$ out of $- x^{2}$ .

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

Step 3.2.1.3

Factor $- 1$ out of $5 x$ .

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

Step 3.2.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$- (x^{2}) - (- 5 x) - 1 \cdot 4 = 0$

Step 3.2.1.5

Factor $- 1$ out of $- (x^{2}) - (- 5 x)$ .

$- (x^{2} - 5 x) - 1 \cdot 4 = 0$

Step 3.2.1.6

Factor $- 1$ out of $- (x^{2} - 5 x) - 1 (4)$ .

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

Step 3.2.2

Factor.

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Step 3.2.2.1

Factor $x^{2} - 5 x + 4$ using the AC method.

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Step 3.2.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1$

Step 3.2.2.1.2

Write the factored form using these integers.

$- ((x - 4) (x - 1)) = 0$

Step 3.2.2.2

Remove unnecessary parentheses.

$- (x - 4) (x - 1) = 0$

Step 3.3

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

$x - 4 = 0$

$x - 1 = 0$

Step 3.4

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

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Step 3.4.1

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

$x - 4 = 0$

Step 3.4.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.5

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

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Step 3.5.1

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

$x - 1 = 0$

Step 3.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.6

The final solution is all the values that make $- (x - 4) (x - 1) = 0$ true.

$x = 4, 1$