Evaluate e^(2x)+e^x-6=0

$e^{2 x} + e^{x} - 6 = 0$

Step 1

Factor the left side of the equation.

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

${(e^{x})}^{2} + e^{x} - 6 = 0$

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

$u^{2} + u - 6 = 0$

Factor $u^{2} + u - 6$ using the AC method.

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

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

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

Step 2

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

$e^{x} - 2 = 0$

$e^{x} + 3 = 0$

Step 3

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

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

$e^{x} - 2 = 0$

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

Add $2$ to both sides of the equation.

$e^{x} = 2$

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

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

Expand the left side.

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

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

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

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

Multiply $x$ by $1$ .

$x = \ln (2)$

Step 4

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

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

$e^{x} + 3 = 0$

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

Subtract $3$ from both sides of the equation.

$e^{x} = - 3$

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

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

The equation cannot be solved because $\ln (- 3)$ is undefined.

Undefined

There is no solution for $e^{x} = - 3$

No solution

Step 5

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

$x = \ln (2)$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (2)$

Decimal Form:

$x = 0.69314718 \dots$

Evaluate e^(2x)+e^x-6=0

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