Precalculus Examples

${(e^{x})}^{x} \cdot e^{45} = e^{14 x}$

Step 1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$e^{x \cdot x} \cdot e^{45} = e^{14 x}$

Step 2

Multiply $e^{x \cdot x}$ by $e^{45}$ by adding the exponents.

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Step 2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$e^{x \cdot x + 45} = e^{14 x}$

Step 2.2

Multiply $x$ by $x$ .

$e^{x^{2} + 45} = e^{14 x}$

Step 3

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x^{2} + 45 = 14 x$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $14 x$ from both sides of the equation.

$x^{2} + 45 - 14 x = 0$

Step 4.2

Factor $x^{2} + 45 - 14 x$ using the AC method.

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Step 4.2.1

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 $45$ and whose sum is $- 14$ .

$- 9, - 5$

Step 4.2.2

Write the factored form using these integers.

$(x - 9) (x - 5) = 0$

Step 4.3

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

$x - 9 = 0$

$x - 5 = 0$

Step 4.4

Set $x - 9$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Step 4.4.2

Add $9$ to both sides of the equation.

$x = 9$

Step 4.5

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 4.5.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 4.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 4.6

The final solution is all the values that make $(x - 9) (x - 5) = 0$ true.

$x = 9, 5$