Precalculus Examples

$e^{2 x} - 6 e^{x} + 9 = 0$

Step 1

Rewrite $e^{2 x}$ as exponentiation.

${(e^{x})}^{2} - 6 e^{x} + 9 = 0$

Step 2

Substitute $u$ for $e^{x}$ .

$u^{2} - 6 u + 9 = 0$

Step 3

Solve for $u$ .

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Step 3.1

Factor using the perfect square rule.

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Step 3.1.1

Rewrite $9$ as $3^{2}$ .

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

Step 3.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Step 3.1.3

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 3 + 3^{2} = 0$

Step 3.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 3$ .

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

Step 3.2

Set the $u - 3$ equal to $0$ .

$u - 3 = 0$

Step 3.3

Add $3$ to both sides of the equation.

$u = 3$

Step 4

Substitute $3$ for $u$ in $u = e^{x}$ .

$3 = e^{x}$

Step 5

Solve $3 = e^{x}$ .

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Step 5.1

Rewrite the equation as $e^{x} = 3$ .

$e^{x} = 3$

Step 5.2

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

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

Step 5.3

Expand the left side.

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

Multiply $x$ by $1$ .

$x = \ln (3)$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (3)$

Decimal Form:

$x = 1.09861228 \dots$