Precalculus Examples

$e^{5 \ln (x - 2)}$

Step 1

Simplify $5 \ln (x - 2)$ by moving $5$ inside the logarithm.

$e^{\ln ({(x - 2)}^{5})}$

Step 2

Exponentiation and log are inverse functions.

${(x - 2)}^{5}$

Step 3

Use the Binomial Theorem.

$x^{5} + 5 x^{4} \cdot - 2 + 10 x^{3} {(- 2)}^{2} + 10 x^{2} {(- 2)}^{3} + 5 x {(- 2)}^{4} + {(- 2)}^{5}$

Step 4

Simplify each term.

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Step 4.1

Multiply $- 2$ by $5$ .

$x^{5} - 10 x^{4} + 10 x^{3} {(- 2)}^{2} + 10 x^{2} {(- 2)}^{3} + 5 x {(- 2)}^{4} + {(- 2)}^{5}$

Step 4.2

Raise $- 2$ to the power of $2$ .

$x^{5} - 10 x^{4} + 10 x^{3} \cdot 4 + 10 x^{2} {(- 2)}^{3} + 5 x {(- 2)}^{4} + {(- 2)}^{5}$

Step 4.3

Multiply $4$ by $10$ .

$x^{5} - 10 x^{4} + 40 x^{3} + 10 x^{2} {(- 2)}^{3} + 5 x {(- 2)}^{4} + {(- 2)}^{5}$

Step 4.4

Raise $- 2$ to the power of $3$ .

$x^{5} - 10 x^{4} + 40 x^{3} + 10 x^{2} \cdot - 8 + 5 x {(- 2)}^{4} + {(- 2)}^{5}$

Step 4.5

Multiply $- 8$ by $10$ .

$x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 5 x {(- 2)}^{4} + {(- 2)}^{5}$

Step 4.6

Raise $- 2$ to the power of $4$ .

$x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 5 x \cdot 16 + {(- 2)}^{5}$

Step 4.7

Multiply $16$ by $5$ .

$x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x + {(- 2)}^{5}$

Step 4.8

Raise $- 2$ to the power of $5$ .

$x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 32$