Precalculus Examples

$e^{4 x} + 2 e^{2 x} - 15 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $e^{4 x}$ as ${(e^{2 x})}^{2}$ .

${(e^{2 x})}^{2} + 2 e^{2 x} - 15 = 0$

Step 1.2

Let $u = e^{2 x}$ . Substitute $u$ for all occurrences of $e^{2 x}$ .

$u^{2} + 2 u - 15 = 0$

Step 1.3

Factor $u^{2} + 2 u - 15$ using the AC method.

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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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 1.3.2

Write the factored form using these integers.

$(u - 3) (u + 5) = 0$

Step 1.4

Replace all occurrences of $u$ with $e^{2 x}$ .

$(e^{2 x} - 3) (e^{2 x} + 5) = 0$

Step 2

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

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

$e^{2 x} + 5 = 0$

Step 3

Set $e^{2 x} - 3$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $e^{2 x} - 3$ equal to $0$ .

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

Step 3.2

Solve $e^{2 x} - 3 = 0$ for $x$ .

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Step 3.2.1

Add $3$ to both sides of the equation.

$e^{2 x} = 3$

Step 3.2.2

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

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

Step 3.2.3

Expand the left side.

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Step 3.2.3.1

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

$2 x \ln (e) = \ln (3)$

Step 3.2.3.2

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

$2 x \cdot 1 = \ln (3)$

Step 3.2.3.3

Multiply $2$ by $1$ .

$2 x = \ln (3)$

Step 3.2.4

Divide each term in $2 x = \ln (3)$ by $2$ and simplify.

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Step 3.2.4.1

Divide each term in $2 x = \ln (3)$ by $2$ .

$\frac{2 x}{2} = \frac{\ln (3)}{2}$

Step 3.2.4.2

Simplify the left side.

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Step 3.2.4.2.1

Cancel the common factor of $2$ .

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Step 3.2.4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\ln (3)}{2}$

Step 3.2.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (3)}{2}$

Step 4

Set $e^{2 x} + 5$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $e^{2 x} + 5$ equal to $0$ .

$e^{2 x} + 5 = 0$

Step 4.2

Solve $e^{2 x} + 5 = 0$ for $x$ .

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Step 4.2.1

Subtract $5$ from both sides of the equation.

$e^{2 x} = - 5$

Step 4.2.2

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

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

Step 4.2.3

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

Undefined

Step 4.2.4

There is no solution for $e^{2 x} = - 5$

No solution

Step 5

The final solution is all the values that make $(e^{2 x} - 3) (e^{2 x} + 5) = 0$ true.

$x = \frac{\ln (3)}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (3)}{2}$

Decimal Form:

$x = 0.54930614 \dots$