Precalculus Examples

$e^{2 x} - e^{x} - 30 = 0$

Step 1

Factor the left side of the equation.

Tap for more steps...

Step 1.1

Rewrite $e^{2 x}$ as ${(e^{x})}^{2}$ .

${(e^{x})}^{2} - e^{x} - 30 = 0$

Step 1.2

Let $u = e^{x}$ . Substitute $u$ for all occurrences of $e^{x}$ .

$u^{2} - u - 30 = 0$

Step 1.3

Factor $u^{2} - u - 30$ using the AC method.

Tap for more steps...

Step 1.3.1

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 $- 30$ and whose sum is $- 1$ .

$- 6, 5$

Step 1.3.2

Write the factored form using these integers.

$(u - 6) (u + 5) = 0$

Step 1.4

Replace all occurrences of $u$ with $e^{x}$ .

$(e^{x} - 6) (e^{x} + 5) = 0$

Step 2

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

$e^{x} - 6 = 0$

$e^{x} + 5 = 0$

Step 3

Set $e^{x} - 6$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.1

Set $e^{x} - 6$ equal to $0$ .

$e^{x} - 6 = 0$

Step 3.2

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

Tap for more steps...

Step 3.2.1

Add $6$ to both sides of the equation.

$e^{x} = 6$

Step 3.2.2

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

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

Step 3.2.3

Expand the left side.

Tap for more steps...

Step 3.2.3.1

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

$x \ln (e) = \ln (6)$

Step 3.2.3.2

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

$x \cdot 1 = \ln (6)$

Step 3.2.3.3

Multiply $x$ by $1$ .

$x = \ln (6)$

Step 4

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

Tap for more steps...

Step 4.1

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

$e^{x} + 5 = 0$

Step 4.2

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

Tap for more steps...

Step 4.2.1

Subtract $5$ from both sides of the equation.

$e^{x} = - 5$

Step 4.2.2

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

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

Step 4.2.3

The equation cannot be solved because $\ln (- 5)$ is undefined.

Undefined

Step 4.2.4

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

No solution

Step 5

The final solution is all the values that make $(e^{x} - 6) (e^{x} + 5) = 0$ true.

$x = \ln (6)$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (6)$

Decimal Form:

$x = 1.79175946 \dots$