Solve for x 2e^(2x)-5e^x-3=0

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

Step 1

Factor the left side of the equation.

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

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

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

$2 u^{2} - 5 u - 3 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 u$ .

$2 u^{2} - 5 u - 3 = 0$

Rewrite $- 5$ as $1$ plus $- 6$

$2 u^{2} + (1 - 6) u - 3 = 0$

Apply the distributive property.

$2 u^{2} + 1 u - 6 u - 3 = 0$

Multiply $u$ by $1$ .

$2 u^{2} + u - 6 u - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} + u) - 6 u - 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (2 u + 1) - 3 (2 u + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

$(2 u + 1) (u - 3) = 0$

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

$(2 e^{x} + 1) (e^{x} - 3) = 0$

Step 2

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

$2 e^{x} + 1 = 0$

$e^{x} - 3 = 0$

Step 3

Set $2 e^{x} + 1$ equal to $0$ and solve for $x$ .

Set $2 e^{x} + 1$ equal to $0$ .

$2 e^{x} + 1 = 0$

Solve $2 e^{x} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 e^{x} = - 1$

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

$\ln (2 e^{x}) = \ln (- 1)$

The equation cannot be solved because $\ln (- 1)$ is undefined.

Undefined

There is no solution for $2 e^{x} = - 1$

No solution

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 $x$ .

Add $3$ to 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.

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

Expand the left side.

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

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

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

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

Multiply $x$ by $1$ .

$x = \ln (3)$

Step 5

The final solution is all the values that make $(2 e^{x} + 1) (e^{x} - 3) = 0$ true.

$x = \ln (3)$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (3)$

Decimal Form:

$x = 1.09861228 \dots$

Solve for x 2e^(2x)-5e^x-3=0

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