Precalculus Examples

$\ln (e^{x^{2} - 3 x}) = 10$

Step 1

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

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Step 1.1

Expand $\ln (e^{x^{2} - 3 x})$ by moving $x^{2} - 3 x$ outside the logarithm.

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

Step 1.2

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

$(x^{2} - 3 x) \cdot 1$

Step 1.3

Multiply $x^{2} - 3 x$ by $1$ .

$x^{2} - 3 x$

Step 2

The expanded equation is $x^{2} - 3 x = 10$ .

$x^{2} - 3 x = 10$

Step 3

Subtract $10$ from both sides of the equation.

$x^{2} - 3 x - 10 = 0$

Step 4

Factor $x^{2} - 3 x - 10$ using the AC method.

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Step 4.1

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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Step 4.2

Write the factored form using these integers.

$(x - 5) (x + 2) = 0$

Step 5

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

$x - 5 = 0$

$x + 2 = 0$

Step 6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 7.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 8

The final solution is all the values that make $(x - 5) (x + 2) = 0$ true.

$x = 5, - 2$