Algebra Examples

$e^{k^{2} + 9} \cdot e^{- k} = e^{2 k + 5} \cdot e^{3 k - 4}$

Step 1

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

$e^{k^{2} + 9 - k} = e^{2 k + 5} \cdot e^{3 k - 4}$

Step 2

Multiply $e^{2 k + 5}$ by $e^{3 k - 4}$ 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.

$e^{k^{2} + 9 - k} = e^{2 k + 5 + 3 k - 4}$

Step 2.2

Add $2 k$ and $3 k$ .

$e^{k^{2} + 9 - k} = e^{5 k + 5 - 4}$

Step 2.3

Subtract $4$ from $5$ .

$e^{k^{2} + 9 - k} = e^{5 k + 1}$

Step 3

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

$k^{2} + 9 - k = 5 k + 1$

Step 4

Solve for $k$ .

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Step 4.1

Move all terms containing $k$ to the left side of the equation.

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Step 4.1.1

Subtract $5 k$ from both sides of the equation.

$k^{2} + 9 - k - 5 k = 1$

Step 4.1.2

Subtract $5 k$ from $- k$ .

$k^{2} + 9 - 6 k = 1$

Step 4.2

Subtract $1$ from both sides of the equation.

$k^{2} + 9 - 6 k - 1 = 0$

Step 4.3

Subtract $1$ from $9$ .

$k^{2} - 6 k + 8 = 0$

Step 4.4

Factor $k^{2} - 6 k + 8$ using the AC method.

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Step 4.4.1

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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 4.4.2

Write the factored form using these integers.

$(k - 4) (k - 2) = 0$

Step 4.5

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

$k - 4 = 0$

$k - 2 = 0$

Step 4.6

Set $k - 4$ equal to $0$ and solve for $k$ .

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Step 4.6.1

Set $k - 4$ equal to $0$ .

$k - 4 = 0$

Step 4.6.2

Add $4$ to both sides of the equation.

$k = 4$

Step 4.7

Set $k - 2$ equal to $0$ and solve for $k$ .

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Step 4.7.1

Set $k - 2$ equal to $0$ .

$k - 2 = 0$

Step 4.7.2

Add $2$ to both sides of the equation.

$k = 2$

Step 4.8

The final solution is all the values that make $(k - 4) (k - 2) = 0$ true.

$k = 4, 2$