Precalculus Examples

$e^{2 x} - 10 e^{x} + 9 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

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

${(e^{x})}^{2} - 10 e^{x} + 9 = 0$

Step 1.2

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

$u^{2} - 10 u + 9 = 0$

Step 1.3

Factor $u^{2} - 10 u + 9$ 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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Step 1.3.2

Write the factored form using these integers.

$(u - 9) (u - 1) = 0$

Step 1.4

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

$(e^{x} - 9) (e^{x} - 1) = 0$

Step 2

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

$e^{x} - 9 = 0$

$e^{x} - 1 = 0$

Step 3

Set $e^{x} - 9$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $e^{x} - 9$ equal to $0$ .

$e^{x} - 9 = 0$

Step 3.2

Solve $e^{x} - 9 = 0$ for $x$ .

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Step 3.2.1

Add $9$ to both sides of the equation.

$e^{x} = 9$

Step 3.2.2

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

$\ln (e^{x}) = \ln (9)$

Step 3.2.3

Expand the left side.

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Step 3.2.3.1

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

$x \ln (e) = \ln (9)$

Step 3.2.3.2

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

$x \cdot 1 = \ln (9)$

Step 3.2.3.3

Multiply $x$ by $1$ .

$x = \ln (9)$

Step 4

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

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Step 4.1

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

$e^{x} - 1 = 0$

Step 4.2

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

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Step 4.2.1

Add $1$ to both sides of the equation.

$e^{x} = 1$

Step 4.2.2

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

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

Step 4.2.3

Expand the left side.

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Step 4.2.3.1

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

$x \ln (e) = \ln (1)$

Step 4.2.3.2

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

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

Step 4.2.3.3

Multiply $x$ by $1$ .

$x = \ln (1)$

Step 4.2.4

The natural logarithm of $1$ is $0$ .

$x = 0$

Step 5

The final solution is all the values that make $(e^{x} - 9) (e^{x} - 1) = 0$ true.

$x = \ln (9), 0$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \ln (9), 0$

Decimal Form:

$x = 2.19722457 \dots, 0$